# How do you solve y = x^2 -8x + 16 graphically and algebraically?

Feb 11, 2018

see below

#### Explanation:

Graphically

the roots are where the graph crosses the $x -$axis.

that is when $y = 0$

graph{x^2-8x+16 [-3.74, 14.04, -2.56, 6.33]}

As can be seen from the graph it touches the $x -$axis at one point only

$x = 4$

Algebraically

we could use factorising., completing the square or the formula.

look for factorising first

${x}^{2} - 8 x = 16 = 0$

${\left(x - 4\right)}^{2} = \left(x - 4\right) \left(x - 4\right) = 0$

$\therefore x - 4 = 0 \implies x = 4$

the repeated brackets show that we have repeated roots, and that the $x -$axis is a tangent to the graph.

Feb 11, 2018

$y = {x}^{2} - 8 x + 16$

Let's solve it algebraically first:

We solve this by finding the "zeros" or x-intercepts of the equation. To do this, we factor the equation, if possible.

We need to find 2 numbers (can be the same) that add up to $- 8$ and multiply to $16$.
$- 4$ works:
$- 4 + - 4 = - 8$
$- 4 \cdot - 4 = 16$

So our equation is:
$y = \left(x - 4\right) \left(x - 4\right)$ or $y = {\left(x - 4\right)}^{2}$

Since we are finding the x-intercept(s), let's set $y = 0$. So:
$0 = {\left(x - 4\right)}^{2}$

To simplify this, let's root both sides by $2$:
$\sqrt{0} = \sqrt{{\left(x - 4\right)}^{2}}$

$0 = x - 4$

$x = 4$

So the x-intercept is at $\left(4 , 0\right)$.

To graph this, we find the vertex and the slope.

In this case, since there is only one x-intercept, that means that it is the vertex.

The slope depends on the coefficient, or the number in front of the highest degree term.
${x}^{2}$ means $1 {x}^{2}$, so our slope is one.

Let's graph this now! As you can see, the x-intercept is the same as the vertex.

Hope this helps!