y = x^2 - 2x - 15...help?

Question

y = x^2 - 2x - 15


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 - 2x - 15
Part a)
y intercept is y = - 15

Part b)
x² - 2x - 15 = 0
(x - 5) (x + 3) = 0
x = - 3 , x = 5 are x intercepts

Part c)
y = (x² - 2x + 1) - 1 - 15
y = (x² - 2x + 1) - 16
y = (x - 1)² - 16
Vertex is (1 , - 16)

Part d)
Line of symmetry is x = 1

Answer 2

a) This is where it crosses the y-axis, so plug in x=0 to get y = -15. The y-intercept is (0,-15).

b) Factor this as y = (x+3)(x-5). Now it should be easier to find the values of x that give y = 0.

c) Use completing the square to get the equation in the form of 4a(y - k) = (x - h)^2. The vertex is (h,k).

d) Since this is a parabola opening upward, the line of symmetry is x = h.

Answer 3

a) just put x=0 in the equation... y = -15 is the y-intercept

b) you must find the roots: delta= b^2-4ac = (-2)^2 -4(1)(-15)
delta = 64 there are two roots
x1= (2+sqrt(64))/2 = 5 and x2 = (2-sqrt(64))/2 = -3

c) the vertex is a point where x = -b/2a , then x = 2/2(1) = 1
to find the y of the vertex put the x in the equation:
y = 1^2 -2(1) - 15 = -16 the vertex is V(1,-16)

d) the line of symmetry is the line y = -b/2a.... then y = 1

Answer 4

a.) let x=0
y-int. is =-15

b.) let y=0, then solve for it

0=x^2-2x-15
0=(x-5)(x+3)
x=5, x= -3 those are your x -int.
c.) vertex (1, -16)

d.) the line of symmetry is at x= 1

Answer 5

Basis:
y = x^2 - 2x - 15
y = (x - 5) (x + 3)

Answers:
a) x = 0, y = - 15
b) y = 0, either x = - 3 or x = 5
c) x = 1, y = - 16
d) x = 1

Answer 6

a) x=0, y= -15
b) y=0, x= -3 or x= 5
c) x=1, y= -16
d) x=1