Need some help with this f(x) equation?

Question

f(x) = 3x^2 + 18x + 24

1. Put this into standard form for a parabola, find the vertex, and find the x and y intercepts.

can someone help me with this problem, thanks!

Answer 1

y=3(x^2+6x+9)-27+24
y=3(x+3)^2-3
y+3=3(x+3)
so the vrtex is (-3,-3)
no y intercept
it is a vertical parabola opening upwads
x intercept(-2,0)and (-4,0)

Answer 2

f(x) = 3x^2 + 18x + 24 [so a=3, b=18, and c=24]
f(x) = 3(x^2 +6x +8) = 3(x+4)(x+2)
The x-intercepts occur when y = f(x) = 0
This occurs when (x+4)= 0 and when ((x+2) = 0
So the x=intercepts are (-4,0) and (-2,0)

The axis of symmetry occurs at x = -b/2a = -18/(2*3) = -3
Since a is positive the parabola is concave downward (like a smile). Thus the minimum value of f(x) occurs when x=-3.
Plug that into f(x) and get f(-3) = 3(-3)^2 =(18)(-3) +24 =
27 -54+24 = -3. So the minimum point of the parabola is (-3,-3) and this is also the vertex of the parabola, since it lies on the axis of symmetry.

Answer 3

-18/2(3)=-3 4(3)(24)-18^2/4(3)=-3
the opening for the parabola is going upward.
the vertex is (-3, -3)
the x- axis is to -2 and -4
no y intercepts

Answer 4

its already in standaed form for a parabola just put it in a graphing calculator. It crosses the x axis at -2 and -4

Answer 5

In response to Raj there has to be a yint always!

yours is (0,24)