Identify the x-intercepts, vertex, and axis of symmetry of the graph of F(x)=-0.25 (x+4)(x+8)

Question

Identify the x-intercepts, vertex, and axis of symmetry of the graph of F(x)=-0.25 (x+4)(x+8)

Answer 1

To find the x-intercepts, set "y" or "f(x)" equal to zero:

0 = -0.25(x + 4)(x + 8)

The only way this product "(x + 4)" multiplied by "(x + 8)" can equal zero is if either "(x + 4)" equals zero or "(x + 8)" equals zero so set each equal to zero and solve:

x + 4 = 0 means x = -4

x + 8 = 0 means x = -8

Therefore the two x-intercepts are x = -4 and x = -8. The points are (-4, 0) and (-8, 0)

The vertex lies halfway in between the two x-intercepts so it must be at x = -6. To know the y-value of the vertex, plug in x = -6:

-0.25((-6) + 4)((-6) + 8) = -0.25(-2)(2) = 1

Therefore the vertex is at (-6, 1)

The axis of symmetry runs right through the vertex and so it is the line x = -6.

Answer 2

F(x)=-0.25 (x+4)(x+8)

The roots for F would correspond to the y-intercept coming out where F(x) = 0, and since F is written in factored form, they are x1= -4 and x2=-8. Obviously y1=y2=0 by definition Since F is a squared polynomial, x for the vertex is x0 = (x1+x2)/2=(-4 - 8)/2=-12/2=-6. the y for the vertex is y0 = F(x0) = F(-6) = -0.25*(-6+4)*(-6+8)=-0.25*(-2)*2=1

In other words,

(x1, y1) = (-4,0) and (x2, y2) = (-8, 0) are two  x-intercepts

vertex is (x0, y0) = (-6, 1), and the axis of symmetry of the graph is x=x0, that is, x=-6

Answer 3

Answer 4

See video.

Answer 5

The graph is a parabola with roots -4 and -8 (i.e. x-intercepts).

The vertex occurs where x=-6 (symmetrically between the roots).

The axis of symmetry is the line x=-6.