what is the axis of symmetry,vertex,and y intercept of the quadratic equation y=x^2-8x+2

Question

finding the axis of symmetry,vertex,and y intercept of the quadratic equation

Answer 1

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

Compare the given equation y = x^2-8x+2 to above equation.

a = 1, b = -8, c = 2

Axis of symmetry is x = -b/2a

x = -(-8)/2(1)

→ x = 4

Vertex of parabola

Substitute the x value in equation y = x^2-8x+2

y = (4)^2-8(4)+2

y = 16-32+2

y = -14

→ Vertex of parabola is (4,-14).

To find y intercept substiute x = 0 in y = x^2-8x+2.

y = (0)^2-8(0)+2

y = 2

→ y intercept is 2.