Y=-x^2-x+20

Question

Please solve for : Finding--- axis of symmetry. X=? Vertex of parabola X intercepts

Answer 1

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

Compare the given equation y = -x^2-x+20 to above equation.

a = -1, b = -1, c = 20

Axis of symmetry is x = -b/2a

x = -(-1)/2(-1)

→ x = -1/2

Vertex of parabola

Substitute the x value in equation y = -x^2-x+20.

y =  -(-1/2)^2-(-1/2)+20

y = -1/4+1/2+20

Least common multiple of 1,2,4 is 4.

y = (-1+2+80)/4

y = 81/4

→ Vertex of parabola is (-1/2,81/4).

To find the x intercept Substitute y = 0 in y = -x^2-x+20.

0 = -x^2-x+20

Multiple to each side by negitive one.

x^2+x-20  = 0

x^2+5x-4x-20 = 0

Now factorize it.

x(x+5)-4(x+5) = 0

(x+5)(x-4) = 0

(x+5) = 0                            and                             (x-4) = 0

Subtract 5 from each side.                         Add 4 to each side.

x+5-5 = 0-5                                                             x-4+4 = 0+4

x = -5  and  x = 4

→ x intercepts are -5,4.