How do you find the x and y intercepts for y=2x^2-12x?

1 Answer
Nov 15, 2015

x = 6,
x = 0.

Explanation:

We start with:
y = 2x^2 -12x.

In order to find the y-intercept, we must substitute x by 0, as the y-axis is x = 0. So,
y = 2(0)^2 - 12(0),
y = 0 - 0,
y = 0.

In order to find the x-intercept, we must substitute y by 0, as the x-axis is y = 0. So,
0 = 2x^2 - 12x,
2(x^2 - 6x) = 0,
x^2 - 6x = 0, according to x = (-b +- sqrt(b^2 -4ac))/(2a),
x = (- (-6) +- sqrt((-6)^2 -4(1)(0))) / (2(1)),
x = (6 +- sqrt(36))/2,
x = (6+6)/2,
x = 12/2,
x = 6,
x = (6-6)/2,
x = 0/2,
x = 0.

Hope it Helps! :D .