Question

find the maximum vertical distance d between the parabola, f(x)=4x^2 + 16x + 6, and the line, g(x)=-x + 2, for the green region.

Explanation:

Step1: Define the distance function

The vertical distance $d$ between the two functions $f(x)=4x^{2}+16x + 6$ and $g(x)=-x + 2$ is given by $d(x)=f(x)-g(x)$. So, $d(x)=4x^{2}+16x + 6-(-x + 2)=4x^{2}+17x + 4$.

Step2: Find the derivative of the distance - function

Differentiate $d(x)$ with respect to $x$. Using the power rule, if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. So, $d^\prime(x)=\frac{d}{dx}(4x^{2}+17x + 4)=8x+17$.

Step3: Find the critical points

Set $d^\prime(x) = 0$ to find the critical points. So, $8x+17 = 0$, which gives $x=-\frac{17}{8}$.

Step4: Determine if it's a maximum or minimum

Differentiate $d^\prime(x)$ to get the second - derivative $d^{\prime\prime}(x)$. $d^{\prime\prime}(x)=\frac{d}{dx}(8x + 17)=8>0$. Since $d^{\prime\prime}(x)>0$, the function $d(x)$ has a minimum at $x =-\frac{17}{8}$. But we can also note that since $y = d(x)$ is a parabola opening upwards ($a = 4>0$ in $y = 4x^{2}+17x + 4$), and we are interested in the region shown (assuming we are looking at the non - infinite part of the green region), we can find the values of $d(x)$ at the intersection points of the parabola and the line.
First, find the intersection points of $f(x)$ and $g(x)$ by setting $4x^{2}+16x + 6=-x + 2$. Rearranging gives $4x^{2}+17x + 4 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$, here $a = 4$, $b = 17$, $c = 4$. So, $x=\frac{-17\pm\sqrt{17^{2}-4\times4\times4}}{2\times4}=\frac{-17\pm\sqrt{289 - 64}}{8}=\frac{-17\pm\sqrt{225}}{8}=\frac{-17\pm15}{8}$. The solutions are $x_1=\frac{-17 + 15}{8}=-\frac{1}{4}$ and $x_2=-4$.

Step5: Evaluate the distance function at the intersection points

$d(x)=4x^{2}+17x + 4$.
When $x =-\frac{1}{4}$, $d(-\frac{1}{4})=4\times(-\frac{1}{4})^{2}+17\times(-\frac{1}{4})+4=4\times\frac{1}{16}-\frac{17}{4}+4=\frac{1}{4}-\frac{17}{4}+4=\frac{1 - 17+16}{4}=0$.
When $x=-4$, $d(-4)=4\times(-4)^{2}+17\times(-4)+4=4\times16-68 + 4=64-68 + 4 = 0$.
We can also note that the vertex of the parabola $y = 4x^{2}+17x + 4$ is at $x=-\frac{17}{8}$.
Let's assume we are looking at the non - negative part of the distance in the green region. We can rewrite $d(x)=4x^{2}+17x + 4$. Completing the square: $d(x)=4(x^{2}+\frac{17}{4}x)+4=4(x^{2}+\frac{17}{4}x+\frac{289}{64}-\frac{289}{64})+4=4((x + \frac{17}{8})^{2}-\frac{289}{64})+4=4(x+\frac{17}{8})^{2}-\frac{289}{16}+4=4(x+\frac{17}{8})^{2}-\frac{289 - 64}{16}=4(x+\frac{17}{8})^{2}-\frac{225}{16}$.
Since the parabola $y = d(x)$ opens upwards, and we are interested in the non - negative part of the distance in the green region, we can find the maximum value of $d(x)$ in the relevant interval.
Let's assume we consider the region between the intersection points.
We can also use the fact that for a quadratic function $y = ax^{2}+bx + c$ ($a>0$), the maximum value in an interval (if not at the vertex) occurs at the endpoints of the interval. But since the distance is $0$ at the intersection points, we made a wrong assumption above. The correct way is to consider the distance formula $d(x)=|f(x)-g(x)|=|4x^{2}+17x + 4|$.
The vertex of the parabola $y = 4x^{2}+17x + 4$ is at $x=-\frac{17}{8}$, and $y = 4\times(-\frac{17}{8})^{2}+17\times(-\frac{17}{8})+4=\frac{4\times289}{64}-\frac{289}{8}+4=\frac{289}{16}-\frac{578}{16}+4=\frac{289 - 578+64}{16}=-\frac{225}{16}$.
The maximum value of $|d(x)|$ in the region occurs at the vertex of $y = 4x^{2}+17x + 4$. So, $d=\frac{225}{16}=14.0625$.

Answer:

$14.0625$