Can you help me!?

Question

1. Find the x- intercept(s) and the coordinates of the vertex for the parabola y = x^2 + 4x - 12. If there is more than one x-intercept seperate them with a comma.


2. What is the roots of this quadratic equation?

t^2 + 5t - 6 = 0

Thank you

Answer 1

1) x-intercepts are precisely the roots of the equation. It factors into (x + 6)(x - 2). The x-intercepts are 2 and -6.
2) Factors nicely as (t - 6)(t + 1). Roots are 6 and -1.

Answer 2

To find the vertex, you mut have your formula in the form y = (x - a)^2 + b. Then the vertex is at (a, b)

You get this by completing the square.

y = x^2 + 4x - 12
. . . 4/2 = 2
. . . 2^2 = 4

y = (x^2 + 4x + 4) - 12 - 4
y = (x + 2)^2 -16

The vertex is where (x + 2) = 0, or x = -2: (-2, -16)

The x-intercepts happen when y = 0.

0 = (x + 2)^2 - 16
x^2 + 4x + 4 - 16 = 0
x^2 + 4x - 12 = 0 (this is your original formula)

x = (-4 +/- sqrt(16 - 4 * 1 * -12)) / (2 * 1)
x = (-4 +/- sqrt(16 + 48)) / 2
x = (-4 +/- sqrt(64)) / 2
x = (-4 +/- 8) / 2
x = 4/2 or -12/2
x = 2 or -6

(2, 0) and (-6, 0)



t^2 + 5t - 6 = 0

t = (-5 +/- sqrt(25 - 4 * 1 * -6)) / (2 * 1)
t = (-5 +/- sqrt(25 + 24)) / 2
t = (-5 +/- sqrt(49)) / 2
t = (-5 +/- 7) / 2
t = -12/2 or 2/2
t = -6 or 1

(t + 6)(t - 1)

Answer 3

1) The easiest way to find roots of a quadratic, is to try to factor (if possible....) In this case it does so easily into:

(x + 6)(x - 2) = 0

giving the roots:

x = -6; x = 2

Factoring is a skill that takes some practice,(like chess....) If the equation cannot be factored easily, then you must use the quadratic formula.....

Finding the min point is easy with the help of a little calculus. Since the slope at the vertex is zero, then I can solve:

d(x² + 4x -12)/dx = 0
2x + 4 = 0
x = -2

If this is over your head, don't worry, just trust me on this one. Suffice it to say, finding zeros and min or max points are some of the most common problems in pure and applied math.

2) (easily) factors into:

(t + 6)(t - 1) = 0

Quandi et demonstranda,
~W.O.M.B.A.T.

Answer 4

Question 1
y = (x² + 4x + 4) - 4 - 12
y = (x + 2)² - 16
(x + 2)² = 16
(x + 2) = ± 4
x = 2 , x = - 6 (intercepts)
Vertex is (- 2, - 16)

Question 2
(t + 6).(t - 1) = 0
t = - 6, t = 1