Question

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

Answer 1

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(x -h)^2 + k`. And if you do this, then the vertex is given by coordinates `(h,k)`.

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

You need to find some number k such that:

`x^2 - 7x + 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 - 7x + 6 + 6.25 - 6.25`

`= x^2- 7x + 12.25 - 6.25`
`= (x - 3.5)^2 - 6.25`
...it's now in vertex form `a(x-h)^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 = (0^2 - 7(0) + 6) = 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 -7x + 6`

gives `x = (7 +- sqrt(-7^2 -4*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]}