Question

How do you graph `y = 3x^2 + 8x - 6`?

Answer 1

Graph `y = 3x^2 + 8x - 6`

Explanation:

To graph a parabola, find first a few critical points. They are: vertex, axis of symmetry, y intercept, and x-intercepts.
a> 0 --> the parabola opens upward.
x- coordinate of vertex and axis of symmetry: `x = -b/(2a) = -8/6 = -4/3`
y-coordinate of vertex:`y = f(-4/3)`=
y intercept --> x = 0 --> y = -6
x-intercepts: solve y = 0
`D = d^2 = b^2 - 4ac = 64 + 72 = 136` --> `d = +- 2sqrt34`
`x = +- -8/6 +- (2sqrt34)/6 = -8/3 +- sqrt34/3`
graph{3x^2 + 8x - 6 [-40, 40, -20, 20]}