1. For \(f(x) = \frac {(x-3)^2}{2} = \frac {1}{2} (x-3)^2+0\), we identify \(a = \frac {1}{2}\), \(h = 3\) and \(k = 0\). Thus the vertex is \((3,0)\) and the parabola opens upwards. The only \(x\)-intercept is \((3,0)\). Since \(f(0) = \frac {1}{2} (0-3)^2 = \frac {9}{2}\), our \(y\)-intercept is \(\left (0, \frac {9}{2}\right )\). To help us graph the function, it would be nice to have a third point and we’ll use symmetry to find it. The \(y\)-value three units to the left of the vertex is \(4.5\), so the \(y\)-value must be \(4.5\) three units to the right of the vertex as well. Hence, we have our third point: \(\left (6, \frac {9}{2}\right )\) and our graph below.

   

   From the graph, we get that the range is \([0, \infty )\) and see that \(f\) has the minimum value of \(0\) at \(x = 3\) and no maximum. Also, \(f\) is decreasing on \((-\infty , 3]\) and increasing on \([3, \infty )\).

2. For \(g(x) = (x+2)^2 - 3 = (1)(x-(-2))^2+(-3)\), we identify \(a = 1\), \(h = -2\) and \(k = -3\). This means that the vertex is \((-2,-3)\) and the parabola opens upwards. Thus we have two \(x\)-intercepts. To find them, we set \(y = g(x) = 0\) and solve. Doing so yields the equation \((x+2)^2 - 3 = 0\), or \((x+2)^2 = 3\). Extracting square roots gives us the two zeros of \(g\): \(x + 2 = \pm \sqrt {3}\), or \(x = -2 \pm \sqrt {3}\). Our \(x\)-intercepts are \((-2-\sqrt {3}, 0) \approx (-3.73, 0)\) and \((-2+\sqrt {3}, 0) \approx (-0.27, 0)\). We find \(g(0) = (0+2)^2-3 = 1\) so our \(y\)-intercept is \((0,1)\). Using symmetry, we get \((-4,1)\) as another point to help us graph.

   

   The range of \(g\) is \([-3, \infty )\). The minimum of \(g\) is \(-3\) at \(x = -2\), and \(g\) has no maximum. Moreover, \(g\) is decreasing on \((-\infty , -2]\) and \(g\) is increasing on \([-2, \infty )\).

3. Given \(h(t) = -2(t-3)^2+1\), we identify \(a = -2\), \(h = 3\) and \(k = 1\). Hence the vertex of the graph is \((3,1)\) and the parabola opens downwards. Solving \(h(t) =-2(t-3)^2+1 = 0\) gives \((t-3)^2 = \frac {1}{2}\). Extracting square roots ^[4] (and rationalizing denominators!) gives \(t - 3 = \pm \frac {\sqrt {2}}{2}\), so that when we add \(3\) to each side, ^[5] (and get common denominators!) we get \(t = \frac {6 \pm \sqrt {2}}{2}\). Hence, our \(t\)-intercepts are \(\left (\frac {6 - \sqrt {2}}{2}, 0 \right ) \approx (2.29, 0)\) and \(\left (\frac {6 + \sqrt {2}}{2}, 0 \right ) \approx (3.71, 0)\). To find the \(y\)-intercept, we compute \(h(0) = -2(0-3)^2+1 = -17\). Thus the \(y\)-intercept is \((0,-17)\). Using symmetry, we also have that \((6,-17)\) is on the graph.

   

   From the graph, we get the range of \(h\) is \((-\infty , 1]\). The maximum of \(h\) is \(1\) at \(x = 3\), and \(h\) has no minimum. We see \(h\) is increasing on \((-\infty , 3]\) and \(h\) is decreasing on \([3, \infty )\).

4. We have some work ahead of us to put \(i(t)\) into a form we can use to exploit Theorem 1:

   \[ \begin{array}{rclcl} i(t) = \frac {(3 - 2t)^2 + 1}{2} & = & \frac {1}{2} (-2t + 3)^2 + \frac {1}{2} & = & \frac {1}{2} \left [ -2 \left (t - \frac {3}{2}\right ) \right ]^2 + \frac {1}{2} \\ & = & \frac {1}{2} (-2)^2 \left (t - \frac {3}{2}\right )^2 + \frac {1}{2} & = & 2\left (t - \frac {3}{2}\right )^2 + \frac {1}{2} \end{array} \]
   We identify \(a = 2\), \(h = \frac {3}{2}\) and \(k = \frac {1}{2}\). Hence our vertex is \(\left (\frac {3}{2}, \frac {1}{2}\right )\) and the parabola opens upwards, meaning there are no \(t\)-intercepts. Since \(i(0) = \frac {(3-2(0))^2 +1}{2} = 5\), we get \((0,5)\) as the \(y\)-intercept. Using symmetry, this means we also have \((3, 5)\) on the graph.
   

   The range is \(\left [ \frac {1}{2}, \infty \right )\) with the minimum of \(i\), \(\frac {1}{2}\), occurring when \(t = \frac {3}{2}\). Also, \(i\) is decreasing on \(\left (-\infty , \frac {3}{2} \right ]\) and increasing on \(\left [\frac {3}{2}, \infty \right )\).

1. We follow the procedure for completing the square in Section ??. The only difference here is instead of the quadratic equation being set to \(0\), it is equal to \(f(x)\). This means when we are finished completing the square, we need to solve for \(f(x)\).

   \[ \begin{array}{rclr} f(x) & = & x^2 - 4x+3 & \\ f(x) - 3 & = & x^2-4x & \text {Subtract $3$ from both sides.} \\ f(x) - 3 + (-2)^2 & = & x^2-4x+(-2)^2 & \text {Add $\left (\frac {1}{2}(-4)\right )^2$ to both sides.} \\ f(x) + 1 & = & (x-2)^2 & \text {Factor the perfect square trinomial.} \\ f(x) & = & (x-2)^2 - 1 & \text {Solve for $f(x)$.} \\ \end{array}\]

   The reader is encouraged to start with \(f(x) = (x-2)^2-1\), perform the indicated operations and simplify the result to \(f(x) = x^2-4x+3\). From the vertex form, \(f(x) = (x-2)^2-1\), we see that the vertex is \((2,1)\) and that the parabola opens upwards. To find the zeros of \(f\), we set \(f(x) = 0\).

   We have two equivalent expressions for \(f(x)\) so we could use either the general form or vertex form. We solve the former and leave it to the reader to solve the latter to see that we get the same results either way. To solve \(x^2 - 4x + 3 = 0\), we factor: \((x-3)(x-1) = 0\) and obtain \(x = 1\) and \(x =3\). We get two \(x\)-intercepts, \((1,0)\) and \((3,0)\).

   To find the \(y\)-intercept, we need \(f(0)\). We use the general form and find that the \(y\)-intercept is \((0,3)\). From symmetry, we know the point \((4,3)\) is also on the graph.

   

   The range of \(f\) is \([-1,\infty )\) with the minimum \(-1\) at \(x = 2\). Finally, \(f\) is decreasing on \((-\infty , 2]\) and increasing from \([2, \infty )\).

2. We rewrite \(g(t) = 6 - 4t - 2t^2\) as \(g(t) = -2t^2 - 4t + 6\) and proceed to complete the square:

   \(g(t)\)	\(=\)	\(-2t^2-4t+6\)
   \(g(t) - 6\)	\(=\)	\(-2t^2-4t\)	Subtract \(6\) from both sides.
   \(\frac {g(t) - 6}{-2}\)	\(=\)	\(\frac { -2t^2-4t }{-2}\)	Divide both sides by \(-2\).
   \(\frac {g(t) - 6}{-2} + (1)^2\)	\(=\)	\(t^2+2t +(1)^2\)	Add \(\left ( \frac {1}{2} (2) \right )^2\) to both sides.
   \(\frac {g(t) - 6}{-2} + 1\)	\(=\)	\((t+1)^2\)	Factor the prefect square trinomial.
   \(\frac {g(t) - 6}{-2}\)	\(=\)	\((t+1)^2 - 1\)
   \(g(t) - 6\)	\(=\)	\(-2 \left [ (t+1)^2-1 \right ]\)
   \(g(t)\)	\(=\)	\(-2(t+1)^2 + 2 + 6\)
   \(g(t)\)	\(=\)	\(-2(t+1)^2+8\)	Solve for \(g(t)\).

   We can check our answer by expanding \(-2(t+1)^2+8\) and show that it simplifies to \(-2t^2 - 4t+6\). From the vertex form, we find that the vertex is \((-1,8)\) and that the parabola opens downwards. Setting \(g(t) = -2t^2 - 4t+6= 0\), we factor to get \(-2(t-1)(t+3) = 0\) so \(t = -3\) and \(t = 1\). Hence, our two \(t\)-intercepts are \((-3,0)\) and \((1,0)\).

   Since \(g(0) = 6\), we get the \(y\)-intercept to be \((0,6)\). Using symmetry, we also have the point \((-2,6)\) on the graph.

   

   The range is \((-\infty , 8]\) with a maximum of \(8\) when \(t = -1\). Finally we note that \(g\) is increasing on \((-\infty , -1]\) and decreasing on \([-1, \infty )\).

1. The formula for the revenue function is \(R(x) = x \, p(x) = x(-1.5x+250) = -1.5x^2 + 250x\). Since the domain of \(p\) is restricted to \(0 \leq x \leq 166\), so is the domain of \(R\). To find the profit function \(P(x)\), we subtract \(P(x) = R(x) - C(x) = \left (-1.5x^2+250x\right ) - \left (80x + 150\right ) = -1.5x^2+170x-150\). The cost function formula is valid for \(x \geq 0\), but the revenue function is valid when \(0 \leq x \leq 166\). Hence, the domain of \(P\) is likewise restricted to \([0, 166]\).
2. We find \(P(0) = -1.5(0)^2+170(0) - 150 = -150\). This means that if we produce and sell \(0\) PortaBoy game systems, we have a profit of \(- \$ 150\). Since \(\text {profit} = \text {(revenue)} - \text {(cost)}\), this means our costs exceed our revenue by \(\$150\). This makes perfect sense, since if we don’t sell any systems, our revenue is \(\$0\) but our fixed costs (see Example ??) are \(\$150\).

3. To find the zeros of \(P\), we set \(P(x) = 0\) and solve \(-1.5x^2+170x-150=0\). Factoring here would be challenging to say the least, so we use the Quadratic Formula, Equation 0.2. Identifying \(a = -1.5\), \(b=170\) and \(c=-150\), we obtain

   \[ \begin{array}{rclr} x & = & \frac {-b \pm \sqrt {b^2-4ac}}{2a} & \\ & = & \frac {-170 \pm \sqrt {170^2 - 4(-1.5)(-150)}}{2(-1.5)} & \\ & = & \frac {-170 \pm \sqrt {28000}}{-3} & \\ & = & \frac {170 \pm 20 \sqrt {70}}{3} & \\ & \approx & 0.89, \, 112.44. \end{array}\]
   Given that \(\text {profit} = \text {(revenue)} - \text {(cost)}\), if \(\text {profit} = 0\), then \(\text {revenue} = \text {cost}\). Hence, the zeros of \(P\) are called the ‘break-even’ points - where just enough product is sold to recover the cost spent to make the product. Also, \(x\) represents a number of game systems, which is a whole number, so instead of using the exact values of the zeros, or even their approximations, we consider \(x = 0\) and \(x = 1\) along with \(x = 112\) and \(x=113\). We find \(P(0) = -150\), \(P(1) = 18.5\), \(P(112) = 74\) and \(P(113)= -93.5\). These data suggest that, in order to be profitable, at least \(1\) but not more than \(112\) systems should be produced and sold, as borne out in the graph below.

4. Knowing the zeros of \(P\), we have two \(x\)-intercepts: \(\left ( \frac {170 - 20 \sqrt {70}}{3},0\right ) \approx (0.89,0)\) and \(\left ( \frac {170 + 20 \sqrt {70}}{3},0\right ) \approx (112.44,0)\). Since \(P(0)=-150\), we get the \(y\)-intercept is \((0,-150)\). To find the vertex, we appeal to Equation 0.1. Substituting \(a = -1.5\) and \(b=170\), we get \(x = -\frac {170}{2(-1.5)} = \frac {170}{3} = 56.\overline {6}\). To find the \(y\)-coordinate of the vertex, we compute \(P\left ( \frac {170}{3} \right ) = \frac {14000}{3} = 4666.\overline {6}\). Hence, the vertex is \((56.\overline {6}, 4666.\overline {6})\). The domain is restricted \(0 \leq x \leq 166\) and we find \(P(166) = -13264\). Attempting to plot all of these points on the same graph to any sort of scale is challenging. Instead, we present a portion of the graph for \(0 \leq x \leq 113\) with the help of Desmos. Even with this, the intercepts near the origin are crowded.

5. From the vertex, we see that the maximum of \(P\) is \(4666.\overline {6}\) when \(x = 56.\overline {6}\). As before, \(x\) represents the number of PortaBoy systems produced and sold, so we cannot produce and sell \(56.\overline {6}\) systems. Hence, by comparing \(P(56) = 4666\) and \(P(57)=4666.5\), we conclude that we will make a maximum profit of \(\$ 4666.50\) if we sell \(57\) game systems.
6. We’ve determined that we need to sell \(57\) PortaBoys to maximize profit, so we substitute \(x=57\) into the price-demand function to get \(p(57) = -1.5(57)+250 = 164.5\). In other words, to sell \(57\) systems, and thereby maximize the profit, we should set the price at \(\$164.50\) per system.

7. To find the average rate of change of \(P\) over \([0, 57]\), we compute

   \[ \frac {\Delta [P(x)]}{\Delta x} = \frac {P(57) - P(0)}{57-0} = \frac {4666.5 - (-150)}{57} = 84.5. \]
   This means that as the number of systems produced and sold ranges from \(0\) to \(57\), the average profit per system is increasing at a rate of \(\$ 84.50\). In other words, for each additional system produced and sold, the profit, on average, increased by \(\$84.50\). □

1. Entering the data in Desmos we find \(T = F(t) = -0.1905t^2+ 3.9643t+62.405\) with an \(R^2\) value of \(0.97\), indicating a pretty strong fit.

2. Since \(7\) AM corresponds to \(t=1\), we find \(T = F(1) \approx 66.18\). Hence our quadratic model predicts a temperature of \(66.2^{\circ }\) F at 7 AM - identical (when rounded) to the \(66.2^{\circ }\) F predicted in Example ??. Similarly, \(3\) PM corresponds to \(t = 9\), so we find \(T = F(9) \approx 82.65\). Thus the model predicts an outdoor temperature of \(82.6^{\circ }\) F which is very close to the \(82.9^{\circ }\)F prediction from Example ?? .

3. The model is quadratic with \(a<0\) so the maximum (warmest) temperature can be determined by finding the vertex. We get

   \[ t = -\frac {b}{2a} = - \frac {3.9643}{2( -0.1905)} \approx -10.40, \quad T = F(-10.40) \approx 83.03,\]
   or, in other words, the warmest temperature is \(83.0^{\circ }\) F at 4:24 PM (\(10.40\) hours after 6 AM.)

   □

Before we dive right into algebraic machinations, let’s take a moment to visualize the problem situation. In the GeoGebra interactive below, we see a representation of the fenced-in pasture on the left. As we move the slider for the width, \(w\), we can see the corresponding shape of the pasture subject to the condition (constraint) that the total fencing available is 200 feet. Below this diagram is a calculator which gives the maximum number of alpaca supported by the pasture. On the right is the graph of the area of the pasture as a function of \(w\) which is traced as we move the slider.

The interactive suggests a width of \(50\) feet will maximize the pasture area allowing Donnie to keep \(200\) alpaca. It’s now time to prove this using the algebraic tools we’ve developed in this section. ^[7] (Mitch Korcheck used to always say that technology suggests solutions; algebra proves them. Rest in peace, Mitch.)

We now have two equations, \(A = w \ell \) and \(\ell +2w = 200\). In order to use the tools given to us in this section to maximize \(A\), we need to use the information given to write \(A\) as a function of just one variable, either \(w\) or \(\ell \). This is where we use the equation \(\ell +2w = 200\). Solving for \(\ell \), we find \(\ell = 200-2w\), and we substitute this into our equation for \(A\). We get \(A = w \ell = w(200-2w) = 200w-2w^2\). We now have \(A\) as a function of \(w\), \(A = f(w) = 200w-2w^2 = -2w^2+200w\).

Before we go any further, we need to find the applied domain of \(f\) so that we know what values of \(w\) make sense in this situation. ^[8] (Donnie would be very upset if, for example, we told him the width of the pasture needs to be \(-50\) feet.) Given that \(w\) represents the width of the pasture we need \(w > 0\). Likewise, \(\ell \) represents the length of the pasture, so \(\ell = 200-2w > 0\). Solving this latter inequality yields \(w < 100\). Hence, the function we wish to maximize is \(f(w) = -2w^2 + 200w\) for \(0 < w < 100\). We know two things about the quadratic function \(f\): the graph of \(A = f(w)\) is a parabola and (since the coefficient of \(w^2\) is \(-2\)) the parabola opens downwards.

This means that there is a maximum value to be found, and we know it occurs at the vertex. Using the vertex formula, we find \(w = -\frac {200}{2(-2)} = 50\), and \(A = f(50) = -2(50)^2 + 200(50) = 5000\). Since \(w=50\) lies in the applied domain, \(0 < w < 100\), we have that the area of the pasture is maximized when the width is \(50\) feet. To find the length, we use \(\ell = 200-2w\) and find \( \ell = 200-2(50) = 100\), so the length of the pasture is \(100\) feet. The maximum area is \(A =f(50) = 5000\), or \(5000\) square feet. If an average alpaca requires 25 square feet of pasture, Donnie can raise \(\frac {5000}{25} = 200\) average alpaca. This solution tracks with what we learned from the GeoGebra interactive earlier. □

1. To solve \(2x^2 \leq 3-x\), we rewrite it as \(2x^2+x-3 \leq 0\). We find the zeros of \(f(x) = 2x^2 + x - 3\) by solving \(2x^2 + x - 3 = 0\). Factoring gives \((2x+3)(x-1)=0\), so \(x = -\frac {3}{2}\) or \(x = 1\). We place these values on the number line with \(0\) above them and choose test values in the intervals \(\left (-\infty , -\frac {3}{2}\right )\), \(\left (-\frac {3}{2},1\right )\) and \((1,\infty )\). For the interval \(\left (-\infty , -\frac {3}{2}\right )\), we choose ^[12] (We have to choose something in each interval. If you don’t like our choices, please feel free to choose different numbers. You’ll get the same sign chart.) \(x=-2\); for \(\left (-\frac {3}{2},1\right )\), we pick \(x=0\); and for \((1,\infty )\), \(x=2\). Evaluating the function at the three test values gives us \(f(-2) = 3 > 0\), so we place \((+)\) above \(\left (-\infty , -\frac {3}{2}\right )\); \(f(0)=-3 < 0\), so \((-)\) goes above the interval \(\left (-\frac {3}{2},1\right )\); and, \(f(2) = 7\), which means \((+)\) is placed above \((1,\infty )\).

   

   We are solving \(2x^2+x-3 \leq 0\) so we need solutions to \(2x^2+x-3 < 0\) as well as solutions for \(2x^2+x-3 =0\). For \(2x^2+x-3 < 0\), we need the intervals which we have a \((-)\) above them. The sign diagram shows only one: \(\left (-\frac {3}{2},1\right )\). Also, we know \(2x^2+x-3 =0\) when \(x=-\frac {3}{2}\) and \(x=1\), so our final answer is \(\left [-\frac {3}{2},1\right ]\).

   To verify our solution graphically, we refer to the original inequality, \(2x^2 \leq 3-x\). We let \(g(x) = 2x^2\) and \(h(x)=3-x\). We are looking for the \(x\) values where the graph of \(g\) is below that of \(h\) (the solution to \(g(x) < h(x)\)) as well as the points of intersection (the solutions to \(g(x)=h(x)\)). The graph below checks our solution is indeed \(\left [-\frac {3}{2},1\right ]\).

   

2. Once again, we re-write \(t^2-2t > 1\) as \(t^2-2t-1>0\) and we identify \(f(t)=t^2-2t-1\). When we go to find the zeros of \(f\), we find, to our chagrin, that the quadratic \(t^2-2t-1\) doesn’t factor nicely. Hence, we resort to the Quadratic Formula and find \(t=1 \pm \sqrt {2}\). As before, these zeros divide the number line into three pieces. To help us decide on test values, we approximate \(1 - \sqrt {2} \approx -0.4\) and \(1 + \sqrt {2} \approx 2.4\). We choose \(t=-1\), \(t=0\) and \(t=3\) as our test values and find \(f(-1)= 2\), which is \((+)\); \(f(0)=-1\) which is \((-)\); and \(f(3)=2\) which is \((+)\) again.

   

   Our solution to \(t^2-2t-1>0\) is where we have \((+)\), so, in interval notation \(\left (-\infty , 1-\sqrt {2}\right ) \cup \left (1+\sqrt {2},\infty \right )\).

   To check the inequality \(t^2 - 2t > 1\) graphically, we set \(g(t) = t^2-2t\) and \(h(t)=1\). We are looking for the \(t\) values where the graph of \(g\) is above the graph of \(h\). The graph below checks our solution is \(\left (-\infty , 1-\sqrt {2}\right ) \cup \left (1+\sqrt {2},\infty \right )\).

   

3. To solve \(x^2+1 \leq 2x\), as before, we solve \(x^2-2x+1 \leq 0\). Setting \(f(x) = x^2-2x+1=0\), we find only one zero of \(f\): \(x = 1\). This one \(x\) value divides the number line into two intervals, from which we choose \(x=0\) and \(x=2\) as test values. We find \(f(0)=1 > 0\) and \(f(2) = 1 > 0\).

   

   Since we are looking for solutions to \(x^2-2x+1 \leq 0\), we are looking for \(x\) values where \(x^2-2x+1 < 0\) as well as where \(x^2-2x+1 = 0\). Looking at our sign diagram, there are no places where \(x^2-2x+1 < 0\) (there are no \((-)\)), so our solution is only \(x=1\) (where \(x^2-2x+1 = 0\)). We write this as \(\left \{1\right \}\).

   Graphically, we solve \(x^2+1 \leq 2x\) by graphing \(g(x) = x^2+1\) and \(h(x)=2x\). We are looking for the \(x\) values where the graph of \(g\) is below the graph of \(h\) (for \(x^2+1 < 2x\)) and where the two graphs intersect (\(x^2+1 = 2x\)). Notice that the line and the parabola touch at \(\left (1, 2\right )\), but the parabola is always above the line otherwise. ^[13] (In this case, we say the line \(y=2x\) is tangent to \(y=x^2+1\) at \(\left (1, 2\right )\). Finding tangent lines to arbitrary functions is a fundamental problem solved, in general, with Calculus.)

   

4. To solve \(2t-t^2 \geq |t-1|-1\) analytically we first rewrite the absolute value using cases. For \(t< 1\), \(|t-1| = -(t-1) = -t+1\), so we get \(2t-t^2 \geq (-t+1)-1\) which simplifies to \(t^2-3t \leq 0\). Finding the zeros of \(f(t) = t^2-3t\), we get \(t=0\) and \(t=3\). However, we are concerned only with the portion of the number line where \(t < 1\), so the only zero that we deal with is \(t=0\). This divides the interval \(t<1\) into two intervals: \((-\infty , 0)\) and \((0,1)\). We choose \(t=-1\) and \(t=\frac {1}{2}\) as our test values. We find \(f(-1) = 4\) and \(f\left (\frac {1}{2}\right ) = -\frac {5}{4}\). Hence, our solution to \(t^2-3t \leq 0\) for \(t < 1\) is \([0,1)\).

   

   Solving \(2t-t^2 \geq |t-1|-1\) for \(t < 1\).

   Next, we turn our attention to the case \(t \geq 1\). Here, \(|t-1| = t-1\), so our original inequality becomes \(2t-t^2 \geq (t-1)-1\), or \(t^2-t-2 \leq 0\). Setting \(g(t) = t^2-t-2\), we find the zeros of \(g\) to be \(t=-1\) and \(t=2\). Of these, only \(t=2\) lies in the region \(t\geq 1\), so we ignore \(t=-1\). Our test intervals are now \([1,2)\) and \((2,\infty )\). We choose \(t=1\) and \(t=3\) as our test values and find \(g(1) = -2\) and \(g(3) = 4\). Hence, our solution to \(g(t) = t^2-t-2 \leq 0\), in this region is \([1,2]\).

   

   Solving \(2t-t^2 \geq |t-1|-1\) for \(t \geq 1\).

   Combining these into one sign diagram, we have that our solution is \([0,2]\).

   

   Graphically, to check \(2t-t^2 \geq |t-1|-1\), we set \(h(t) = 2t-t^2\) and \(i(t) = |t-1|-1\) and look for the \(t\) values where the graph of \(h\) intersects or is above the the graph of \(i\). Our graph below checks that our answer is \([0,2]\).

   

Next, we graph \(f(x) = x^2-x-6\).

If we take a step back and look at the graphs of \(f\) and \(g\), we notice that to obtain the graph of \(g\) from the graph of \(f\), we reflect the portion of the graph of \(f\) which lies below the \(x\)-axis to above the \(x\)-axis and leave the portion of the graph of \(f\) which lies above the \(x\)-axis alone.

In general, if \(g(x) = |f(x)|\), then: