Question

Find the vertex, axis of symmetry
y=4x^2-16x+3
Find the vertex, axis of symmetry
y=4x^2-16x+3
@Mathematics

Answer 1

For the vertex,

x - value of vertex = -b / 2a
= -(-16) / 2(4)
= 2

y -value of vertex = 4(2)^2 - 16(2) + 3
= 16 - 32 + 3
= -13

vertex is at (2, -13)

Answer 2

Axis of symmetry is x = 2

Answer 3

what would be the x and y intercepts?

Answer 4

x intercepts are the zeroes.

x = 3.8 and x = 0.19

The y-intercept is when x = 0.

y = 4(0)^2 - 16(0) + 3
= 0 - 0 +3
= 3

Answer 5

You can round 0.19 to 0.2.