Problema Solution
For the function f(x) = 2x^2 + 8x - 24
What is the Vertex? What is the Axis of Symmetry? Describe the max or min value of the function. What is the y-intercept? What equation would you solve to find the x-intercepts of the function.
Answer provided by our tutors
f(x) = 2x^2 + 8x - 24
let's define: a=2, b=8 and c=-24;
VERTEX: x=-b/2a = -8/(2*2)= -8/4 = -2
VERTEX: y= 2x^2 + 8x - 24= 2*(-2)^2 +8*(-2) - 24= 2*4 - 16 - 24 = -32
so VERTEX has coords: (-2,-32);
We have an axe of simmetry y where there is the x coord of the vertex so: SIMMETRY: x=-2;
Consider the derivate of f(x)=f'(x)=4x+8, so special points are where f'(x)=0:
4x+8=0 -> 4x=-8 -> x=-2 --> so in P(-2,-32) is a point where function assumes the min value because concavity is up side.
The funcion intercept the y axe in c --> so in y=-24
To see the x-intercepts of the function we must solve:
x1 = [-b+sqrt(b^2-4ac)]/2a --> [-8+sqrt(64+192)]/4 --> (-8+16)/4 --> x1=2
x2 = [-b-sqrt(b^2-4ac)]/2a --> [-8-sqrt(64+192)]/4 --> (-8-16)/4 --> x2=-6