quadratic equations, anyone good at them?

Question

y = - x^2 - 4x + 12

For the above quadratic equation find:

a.The y-intercept.

b.The x-intercepts if possible.

c.The vertex.

d. The line of symmetry.

Answer 1

y = - x² - 4x + 12

Part a)
y intercept is y = 12

Part b)
-x² - 4x + 12 = 0
x² + 4x - 12 = 0
(x + 6) (x - 2) = 0
x = - 6 , x = 2 are x intercepts

Part c)
y = - (x² + 4x - 12)
y = - (x² + 4x + 4 - 4 - 12)
y = - (x² + 4x + 4 - 16)
y = - (x² + 4x + 4) - 16
y = - (x + 2)² - 16
Vertex is (- 2 , - 16)

Part d)
Axis of symmetry is x = - 2

Answer 2

a) Set x = 0 and you have y = 12

b) Set y = 0 for the roots and with the quadratic formula you have
(4+-sqrt(16+4*12))/(-2) = -6, 2

c) Since the roots are -6 and 2, the curve must turn on either side of the x-axis at the midpoint between these roots, so
(-6+2)/2 = -2 for x. Substitute in the equation and it yields
y = 12 and the vertex is at (-2, 12).

d) The axis of symmetry is the equation of the line passing through the vertex or x = -2.

Answer 3

a. The y-intercept occurs when x=0, so plug 0 into the equation.

y=-(0)^2+4(0)+12
=12

b. To get the x-intercepts (also called the zeros, or roots), you will need to factor it.

y=-x^2-4x+12
=(x-6)(x+2)

The x-intercepts are therefore 6, and -2

c. Complete the square for the vertex.

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

Therefore, the vertex is at (-2,16)

d. The line of symmetry is half of the sum of the two zeros.

1/2(6-2)

=2

The line of symmetry is at x=2

Answer 4

To find the y-intercept, substitute x=0. So y=12.

To find the x intercepts, factorise or use the general formula. In this case, factorisation is possible, so y= (6+x)(2-x)
Substitute y=0, x=-6, and x=2

To find the vertex, complete the square. y=-(x+2)^2 +16
Vertex= (-2,16)

The line of symmetry is y=-2

Answer 5

we can write it in the form y = - (x+2)^2 +16 then to graph it move to the left 2units move up 16 units then cup down parabola. then
a.(0,12)
b.(2,0) and (-6,0)
c.(- 2,16)
d. x= - 2