# How do you graph the parabola y=x^2-7x+6 using vertex, intercepts and additional points?

Sep 1, 2017

I had to go back and refresh my memory on how to do this.

Anyway, what you have to do is, rewrite the initial equation in what is called "vertex form":

$y = a {\left(x - h\right)}^{2} + k$. And if you do this, then the vertex is given by coordinates $\left(h , k\right)$.

So, this is your infamous "complete the square" operation.

You need to find some number k such that:

${x}^{2} - 7 x + k$ is a perfect square.

Well, if you divide -7 by 2, then square this, you get $- {3.5}^{2} = 12. 25$

So, if you add 6.25 to 6, you'll have a perfect square.

It's an algebraic trick, but you can write:

$y = {x}^{2} - 7 x + 6 + 6.25 - 6.25$

$= {x}^{2} - 7 x + 12.25 - 6.25$
$= {\left(x - 3.5\right)}^{2} - 6.25$
...it's now in vertex form $a {\left(x - h\right)}^{2} + k$, with a = 1, h = 3.5, k = -6.5

Your vertex is at point (3.5, -6.5)

To find the y intercept, set x = 0 and solve for y. I'll use the original form of the equation to calculate:

$y = \left({0}^{2} - 7 \left(0\right) + 6\right) = 6$

x intercept is found by setting y = 0 and solving for x. This is just finding the roots of this quadratic equation.

$0 = {x}^{2} - 7 x + 6$

gives $x = \frac{7 \pm \sqrt{- {7}^{2} - 4 \cdot 6}}{2}$

giving roots 6 and 1.

To graph the curve, plot points (3.5, -6.5), (6,0), and (1, 0).

You can plot any additional points you want. Say, x = 2:

y = 4 - 14 + 6 = -4.

Connect the dots with a smooth, pretty hyperbola, and bam:

graph{x^2 - 7x + 6 [-9.45, 15.36, -8.83, 3.57]}