Quadratic graph with vertex at $(3,-5)$ and $y$-intercept $(0,31)$

Expert intro: This question combines key ideas about quadratic graphs: using vertex form to recover coefficients, then applying a vector translation to the equation. The main goal is to connect geometric features of the parabola with its algebraic equation.

Detailed walkthrough

(a) Find $a$, $b$ and $c$

Since the vertex is $(3,-5)$, write the quadratic in vertex form:

$y=a(x-3)^2-5$

The graph intersects the $y$-axis at $(0,31)$, so substitute $x=0$, $y=31$:

$31=a(0-3)^2-5$ $31=9a-5$ $9a=36$ $a=4$

Now expand:

$y=4(x-3)^2-5$ $y=4(x^2-6x+9)-5$ $y=4x^2-24x+36-5$ $y=4x^2-24x+31$

So:

• $a=4$
• $b=-24$
• $c=31$

(b) Translate by $\begin{pmatrix}-2\\7\end{pmatrix}$

A translation by $(-2,7)$ means:

• shift the graph 2 units left,
• shift the graph 7 units up.

For $y=f(x)$, the translated graph is:

$y=f(x+2)+7$

Using $f(x)=4x^2-24x+31$:

$y=4(x+2)^2-24(x+2)+31+7$

Simplify:

$y=4(x^2+4x+4)-24x-48+38$ $y=4x^2+16x+16-24x-48+38$ $y=4x^2-8x+6$

So the equation of the translated graph is:

$\boxed{y=4x^2-8x+6}$

💡 Pitfall guide

A common mistake is to use the vertex directly as if the equation were $y=a(x+3)^2-5$. The correct sign is $y=a(x-3)^2-5$ because the vertex is at $x=3$. Another frequent error is forgetting that a translation by $\begin{pmatrix}-2\\7\end{pmatrix}$ means left 2 and up 7, so the new equation is $y=f(x+2)+7$ rather than $y=f(x-2)+7$.

🔄 Real-world variant

If the translation vector were $\begin{pmatrix}p\\q\end{pmatrix}$ instead, the transformed graph of $y=f(x)$ would be

$y=f(x-p)+q$

So for this parabola, any horizontal or vertical shift can be handled directly from the original equation $y=4x^2-24x+31$.

🔍 Related terms

vertex form, quadratic transformation, translation vector