Finding components of a quadratic function.

Question

The following diagram shows part of the graph of a quadratic function f.

The vertex is at $(3, -1)$ and the x intercepts are at $2$ and $4$.

The function f can be written in the form $f(x) = (x - h)^2 + k$.

a. Write down the value of $h$ and of $k$.

Aren’t $h$ and $k$ the vertex? So, $h = 3$ and $k= -1$?

b. The function can also be written in the form $f(x) = (x - a) (x - b)$.

Write down the value of $a$ and $b$.

Would we plug in the two $x$-intercepts, $2$ and $4$, in this function and find a and $b$?

c. Find the $y$-intercept.

If we use the $x$ intercepts and find $a$ and $b$ in the last part , couldn’t we solve it and that would give us our $y$?

Answer 1

Aren’t $h$ and $k$ the vertex? So, $h=3$ and $k=−1$?

Yeah, that's right. Because if $f = (x - h)^2 + k$, then $f$ is minimal iff $(x-h)^2$ is minimal, which is minimal iff $x - h = 0$, i.e. $x = h$. So we see that $h$ is the $x$-coordinate of the vertex. Plugging this into $f(x) = (x - h)^2 + k$ we see that $f(h) = k$. Thus $(h, k)$ is the vertex.

Would we plug in the two $x$-intercepts, $2$ and $4$, in this function and find $a$ and $b$?

Almost right. If $f(x) = (x-a)(x-b)$, then $f(x) = 0$ iff $x = a$ or $x = b$, as one of the factors has to be zero. So we see that $a$ and $b$ actually are the $x$-intercepts of $f$.

If we use the x intercepts and find $a$ and $b$ in the last part , couldn’t we solve it and that would give us our y?

Yep. You just have to plug $x=0$ into your $f$, i.e. calculate $f(0)$.