Write f(x)=1/2x^2+1/2x−6 in intercept form. Then find the x-intercepts, the vertex, and the axis of symmetry of the graph of f.

Question

Once Again Need Help...............

Answer 1

To find intercept form, we need to factor our equation.

1. First, factor out the first coefficient -> f(x) = (1/2)(x² + x - 12)
2. Second, factor the quadratic
3. We need two numbers that multiply to -12 (last term) and add to +1 (middle term)
4. These two numbers, in this case are 4 and -3
5. 4 + -3 = 1
6. 4 * -3 = -12
7. Therefore, x² + x - 12 = (x + 4)(x - 3)
8. Intercept form -> f(x) = (1/2)(x+4)(x-3)
9. The x-intercepts will be where the factors = 0
10. x + 4 = 0 -> x = -4
11. x - 3 = 0 -> x = 3
12. Therefore, the x-intercepts are x=-4 and x=3
13. To find the axis of symmetry, we simply take the average of our intercepts (to find the center of our parabola)
14. (-4 + 3)/2 = -1/2
15. Axis of symmetry -> x = -1/2
16. To find the vertex, we plug the axis of symmetry into our equation
17. f(x) = (1/2)(x + 4)(x - 3) = (1/2)(-1/2 + 4)(-1/2 - 3) = (1/2)(7/2)(-7/2) = -49/8
18. vertex -> (-1/2, -49/8)

Answer 2

f(x) = 1/2x² + 1/2x - 6

f(x) = (1/2x + 2)(x - 3)

x-intercepts:

1/2x + 2 = 0

1/2x = -2

x = -4

x - 3 = 0

x = 3

vertex:

x = -b/2a

x = -1/2/(2*1/2) = -1/2

f(1/2) = (1/2(1/2) + 2)(1/2 -3)

f(1/2) = 2.25 * -2.5 = -5.625

(-0.5,-5.625)

axis of symmetry:

x = -0.5