what is the axis of summetry and vertex and yint and xint of -x^2+2x+15

Question

Help m e pleaseee, i dont know how to do thisssss!!!!!

Answer 1

Standard equation of parabola is y = ax^2+bx+c

Compare the given equation y = -x^2+2x+15 to above equation.

a = -1, b = 2, c = 15

Axis of symmetry is x = -b/2a

x = -2/2(-1)

→ x = 1

Vertex of parabola

Substitute the x value in equation y = -x^2+2x+15

y =  -(1)^2+2(1)+15

y = -1+2+15

y = 16

→ Vertex of parabola is (1,16).

To find the x intercept Substitute y = 0 in y = -x^2+2x+15

0 = -x^2+2x+15

Multiple to each side by negitive one

x^2-2x-15 = 0

Now factorize it.

x^2-5x+3x-15 = 0

x(x-5)+3(x-5) = 0

(x-5)+(x+3) = 0

(x-5) = 0                             and            (x+3) = 0

Add 5 to each side.                   subtract 3 from each side.

x-5+5 = 0+5                                   x+3-3 = 0-3

x = 5   and x = -3   

→ x intercepts are 5,-3.

To find the y intercept Substitute x = 0 in y = -x^2+2x+15.

y = -(0)^2+2(0)+15

y = 15

→ y intercept is 15.