12) The x-intercept is: x = 4 or -2
The y-intercept is: y = -8
13) The equation in standard form is: y = x² - 2x - 8
14) The vertex will be at the point: (1, -9)
15) The graph is plotted and attached
How to solve the quadratic equation?
The equation of the parabola is given as:
y = (x - 4)(x + 2)
12) The x-intercept will be the point where y = 0. Thus:
(x - 4)(x + 2) = 0
x = 4 or -2
The y-intercept is the point where x = 0. Thus:
y = (0 - 4)(0 + 2)
y = -8
13) Expanding the equation in standard form is:
y = (x - 4)(x + 2)
y = x² - 2x - 8
14) The vertex will be at the point:
x = -b/2a
x = -(-2)/2(1)
x = 1
At x = 1:
y = 1² - 2(1) - 8
y = -9
Vertex is at (1, -9)
15) The graph is plotted and attached