Question

write an equation in standard form of the parabola that has the same shape as the graph of f(x) = 4x^2, but with (9,1) as the vertex. g(x)= (type your answer in standard form.)

Explanation:

Step1: Recall the vertex - form of a parabola

The vertex - form of a parabola is $y = a(x - h)^2+k$, where $(h,k)$ is the vertex of the parabola. Since the parabola $g(x)$ has the same shape as $f(x)=4x^{2}$, the value of $a$ is the same as the coefficient of $x^{2}$ in $f(x)$, so $a = 4$. The vertex of $g(x)$ is given as $(h,k)=(9,1)$.

Step2: Substitute the values of $a$, $h$, and $k$ into the vertex - form

Substitute $a = 4$, $h = 9$, and $k = 1$ into the equation $y=a(x - h)^2+k$. We get $g(x)=4(x - 9)^2+1$.

Step3: Expand the equation to standard form

Expand $4(x - 9)^2+1$:
\[

$$\begin{align*} 4(x - 9)^2+1&=4(x^{2}-18x + 81)+1\\ &=4x^{2}-72x+324 + 1\\ &=4x^{2}-72x+325 \end{align*}$$

\]

Answer:

$g(x)=4x^{2}-72x + 325$