Question

an engineer is designing a parabolic dish to receive radio signals. she graphs her design on a coordinate plane to show the details of the dish. the graph shows the height, y, of the dish, in inches, in terms of x, the diameter in inches. which equation represents this graph? a. $y = 10(x^{2}-40x + 300)$ b. $y=\frac{1}{10}(x^{2}-40x + 300)$ c. $y = 10(x^{2}-4x + 30)$ d. $y=\frac{1}{10}(x^{2}-4x + 30)$

Explanation:

Step1: Recall vertex - form of parabola

The general form of a quadratic function is $y = ax^{2}+bx + c$. The vertex - form is $y=a(x - h)^{2}+k$, where $(h,k)$ is the vertex of the parabola. From the graph, the vertex of the parabola is $(20, - 10)$. So the equation of the parabola can be written as $y=a(x - 20)^{2}-10$.

Step2: Use a point on the parabola

The parabola passes through the point $(0,30)$. Substitute $x = 0$ and $y = 30$ into $y=a(x - 20)^{2}-10$.
We get $30=a(0 - 20)^{2}-10$.

Step3: Solve for $a$

First, simplify the equation:
\[

$$\begin{align*} 30&=a\times400-10\\ 30 + 10&=400a\\ 40&=400a\\ a&=\frac{1}{10} \end{align*}$$

\]
Then expand $y=\frac{1}{10}(x - 20)^{2}-10$.
\[

$$\begin{align*} y&=\frac{1}{10}(x^{2}-40x + 400)-10\\ y&=\frac{1}{10}x^{2}-4x + 40-10\\ y&=\frac{1}{10}(x^{2}-40x + 300) \end{align*}$$

\]

Answer:

B. $y=\frac{1}{10}(x^{2}-40x + 300)$