Question

What is the axis of symmetry, vertex and y intercept for 2x^2-8x-5

Answer 1

Answer:

Axis of symmetry: x = 2

Vertex at: (2, -13)

y-intercept is (0, -5)

Step-by-step explanation:

The equation for the x-value of the vertex in a quadratic of the form:

$y=ax^2+bx+c$

is given by:

$x_{vert}=-\frac{b}{2\,a}$

which in our case gives:

$x_{vert}=\frac{8}{2*2} =2$

Then the equation for the axis of symmetry is given by:  x = 2

and the y of the vertex can be calculated as:

$y_{vert}=2 (2)^2-8\,(2)-5=-13$

Then the vertex is at the point: (2, -13)

And the y-intercept is found when x=0, that is:

y-intercept is (0, -5)