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

Question

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

Ethan Adkins · Accepted Answer

Vertex (1, -11)
#x = -1 +- sqrt3/3# x-coordinate of the vertex: #x = -b/(2a) = 6/-6 = 1# y-coordinate of vertex: #y(1) = - 3 - 6 - 2 = -11# Vertex (1, -11) Make x = 0, y-intercept = - 2 To find x-intercepts, make y = 0 and solve the quadratic equation: #f(x) = -3x^2 - 6x -2 = 0.# #D = d^2 = b^2 - 4ac = 36 - 24 = 12# --> #d = +- 2sqrt3# There are 2 x-intercepts (real roots): #x = -b/(2a) +- d/(2a) = 6/-6 +- 2sqrt3/6 = -1 +- sqrt3/3#

Jameson Allen · Answer

undefined To find the vertex of the quadratic function $ f(x) = -3x^2 - 6x - 2 $, you first need to determine the x-coordinate of the vertex, which is given by $ x = -\frac{b}{2a} $, where $ a $ and $ b $ are the coefficients of the quadratic equation. In this case, $ a = -3 $ and $ b = -6 $. Then, substitute these values into the formula to find $ x $.

To find the y-coordinate of the vertex, plug the $ x $-coordinate found in the previous step into the original function $ f(x) $.

To find the x-intercepts (or roots), set $ f(x) = 0 $ and solve for $ x $. You can use the quadratic formula or factor the quadratic equation if possible.

To find the y-intercept, simply plug in $ x = 0 $ into the function $ f(x) $.