Find the -intercept(s) and the coordinates of the vertex for the parabola

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

the coordinates of parabola vertex

y = x² + 4x -12
y = (x² + 4x + 4) - 16
y = (x + 2)² - 16

∴ the coordinates is (-2 , -16)

the intercepts:


for y = 0
x² + 4x -12 = 0
(x + 6)(x - 2) = 0
x = 2 and -6

for x = 0
y = -12

∴ the interceps (0,-12) (2,0) (-6,0)

The vertex is when the derivative of f(x)=ax^2+bx+c is 0

f'(x)=2ax+b=0
2ax=-b
x=-b/2a

f(-b/2a) = a(-b/2a)^2 + b(-b/2a) + c
f(-b/2a) = a(b^2/4a^2) -b^2/2a + c
f(-b/2a) = (b^2/4a) -b^2/2a + c
f(-b/2a) = b^2/4a -2b^2/4a + c
f(-b/2a) = -b^2/4a + c
f(-b/2a) = c-b^2/4a

So the vertex is at (-b/2a, c-b^2/4a).


The x-intercepts are when in f(x)=ax^2+bx+c, f(x)=0
ax^2+bx+c=0
x^2+(b/a)x = -c/a
x^2+(b/a)x + b^2/4a^2 = -c/a + b^2/4a^2
(x+b/2a)^2 = -c/a + b^2/4a^2
(x+b/2a)^2 = -c/a(4a/4a) + b^2/4a^2
(x+b/2a)^2 = (-4ac)/(4a^2) + b^2/4a^2
(x+b/2a)^2 = (b^2-4ac)/(4a^2)
x+b/2a = +/-sqrt[b^2-4ac]/(2a)
x = -b/2a +/-sqrt[b^2-4ac]/(2a)
x = [-b+/-sqrt(b^2-4ac)]/(2a)

Now that I've derived everything for you, do it yourself.

well, the y-intercept occurs when x=0, and the x-intercept occurs when y=0.

so if we have y = x^2 + 4x - 12,

when x = 0, y= -12. therefore the y-intercept is y= -12.

when y= 0:

0 = x^2 + 4x - 12

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

x = -6, 2

the x-intercepts are x = -6 and x=2