What are the x-intercepts of the parabola x2 + 4x −12?
I figured it out
x-intercept occurs at y=0. thus we substitute it to the given equation,
y = x^2 + 4x - 12
0 = x^2 + 4x - 12
0 = (x-2)(x+6)
x = 2 and x = -6
at these values of x, y is zero. thus the x-intercepts are
(2,0) and (-6,0)
hope this helps~ :)
To find the x-intercepts of a parabola, you need to find the values of x when y (the value of the function) is equal to zero.
In this case, we have the equation x^2 + 4x - 12 = 0. To solve for x, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the equation.
From the equation x^2 + 4x - 12 = 0, we have a = 1, b = 4, and c = -12. Substituting these values into the quadratic formula, we get:
x = (-4 ± √(4^2 - 4(1)(-12))) / (2(1))
= (-4 ± √(16 + 48)) / 2
= (-4 ± √64) / 2
= (-4 ± 8) / 2
Now we can find the x-intercepts by evaluating both possible solutions:
x1 = (-4 + 8) / 2 = 4 / 2 = 2
x2 = (-4 - 8) / 2 = -12 / 2 = -6
Therefore, the x-intercepts of the parabola x^2 + 4x - 12 are 2 and -6.