Question

How do you find the vertex and the intercepts for `f(x)= -3x^2-6x-7`?

Answer 1

Vertex (-1, -4)
y-intercept = - 7

Explanation:

`f(x) = - 3x^2 - 6x - 7`
x-coordinate of vertex:
`x = - b/(2a) = 6/-6 = - 1`
y-coordinate of vertex:
f(-1) = - 3 + 6 - 7 = -4
Vertex (-1, -4)
To find the 2 x-intercepts, solve the quadratic equation:
`f(x) = - 3x^2 - 6x - 7 = 0`
`D = d^2 = b^2 - 4ac = 36 - 84 = - 48 < 0`
There are no x-intercepts (no real roots) because D < 0.
To find y-intercept, make x = 0
y-intercept = - 7.
Since a < 0, the parabola graph opens downward and stays completely below the x-axis.
graph{- 3x^2 - 6x - 7 [-20, 20, -10, 10]}