How do you find the vertex and intercepts for f(x)=x^2-16x+63?

Mar 26, 2018

$y$-intercept: $y = 63$
$x$-intercepts: $x = 7$ and $x = 9$
Vertex" $\left(x , y\right) = \left(8 , 1\right)$

Explanation:

Intercepts
The $y$ (also know as $f \left(x\right)$) intercept is the value of $y$ when $x = 0$
Given $y = f \left(\textcolor{b l u e}{x}\right) = {\textcolor{b l u e}{x}}^{2} - 16 \textcolor{b l u e}{x} + 63$
when $\textcolor{b l u e}{x} = \textcolor{b l u e}{0}$
$\textcolor{w h i t e}{\text{XXX}} y = f \left(\textcolor{b l u e}{0}\right) = {\textcolor{b l u e}{0}}^{2} - 16 \cdot \textcolor{b l u e}{0} + 63 = \textcolor{red}{63}$
So the $y$ intercept is at $y = 63$
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[Depending upon prior knowledge, the following could be greatly reduced; however, I decided to provide the complete detailed expansion].
The $x$-intercepts are the values of $\textcolor{b l u e}{x}$ for which $f \left(\textcolor{b l u e}{x}\right) = \textcolor{b l u e}{0}$
$\textcolor{w h i t e}{\text{XXX}} f \left(\textcolor{b l u e}{x}\right) = \textcolor{b l u e}{0} = {x}^{2} - 16 x + 63$
We know that if $f \left(x\right)$ can be factored into two binomials of the form:
$\textcolor{w h i t e}{\text{XXX}} \left(x + \textcolor{m a \ge n t a}{p}\right) \left(x + \textcolor{\lim e}{q}\right)$
then by expansion
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = {x}^{2} + \left(\textcolor{m a \ge n t a}{p} + \textcolor{\lim e}{q}\right) x + \textcolor{m a \ge n t a}{p} \textcolor{\lim e}{q}$

We are given that
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = {x}^{2} - 16 x + 63$
So if $f \left(x\right)$ can be factored into the form $\left(x + \textcolor{m a \ge n t a}{p}\right) \left(x + \textcolor{\lim e}{q}\right)$
then
$\textcolor{w h i t e}{\text{XXX}} \textcolor{m a \ge n t a}{p} \textcolor{m a \ge n t a}{q} = 63$
and
$\textcolor{w h i t e}{\text{XXX}} \textcolor{m a \ge n t a}{p} + \textcolor{\lim e}{q} = - 16$

That is, we are hoping to find factors of $63$ which add up to $\left(- 16\right)$.
Because $\textcolor{m a \ge n t a}{p} \textcolor{\lim e}{q} = 63 > 0$ we know that $\textcolor{m a \ge n t a}{p}$ and $\textcolor{\lim e}{q}$ must have the same sign;
Furthermore, since $\textcolor{m a \ge n t a}{p} + \textcolor{\lim e}{q} = - 16$, both $\textcolor{m a \ge n t a}{p}$ and color(lime)q must be negative.

We can start building a list of such possible factors:
color(white)("XXX"){: (color(white)("xx")color(magenta)p,color(white)("xx")color(lime)q,color(white)("XXX"),"Sum"), (-1,-63,,-64), (-3,-21,,-24), (-7,-9,,-16) :}
There is no need to continue beyond this point since we have found a pair of values $\textcolor{m a \ge n t a}{p}$ and color(lime)q that meet our requirements.

We now can write
$\textcolor{w h i t e}{\text{XXX}} 0 = \left(x \textcolor{m a \ge n t a}{- 7}\right) \left(x \textcolor{\lim e}{- 9}\right)$
which implies
either$\textcolor{w h i t e}{\text{XXX}} x = 7$ or $\textcolor{w h i t e}{\text{XXX}} x = 9$

That is the x-intercepts are at $x = 7$ and $x = 9$
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Vertex
$f \left(x\right) = {x}^{2} - 16 x + 63$
can be converted into vertex form $f \left(x\right) = {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$, as follows
$f \left(x\right) = {x}^{2} - 16 x \textcolor{\mathmr{and} a n \ge}{+ {8}^{2}} + 63 \textcolor{\mathmr{and} a n \ge}{- {8}^{2}}$
$\textcolor{w h i t e}{\text{XXX}} = \left(x - \textcolor{red}{8}\right) + \textcolor{b l u e}{1}$

That is, the vertex is at $\left(x , y\right) = \left(8 , 1\right)$