How do you find the equation of the axis of symmetry and the coordinates of the vertex of the graph of the function y= -2x^2+6x-1

Question

asked Oct 6, 2020 8.7k views

1 Answer

Dpdwilson · Answer 1 · 2020-10-12T13:53:58+0000

Answer:

The vertex is:
$((3)/(2),\ (7)/(2))$

The axis of symmetry is:

x=(3)/(2)

Explanation:

For a quadratic equation of the form:

y=ax^2 + bx +c

The vertex of the parabola will be the point:
$(-(b)/(2a),\ f(-(b)/(2a)))$

In this case we have the following equation:

y= -2x^2+6x-1

Note that:

$a=-2\\b=6\\c=-1$

Then the x coordinate of the vertex is:

x=-(b)/(2a)

x=-(6)/(2(-2))

x=(3)/(2)

Then the y coordinate of the vertex is:

y= -2((3)/(2))^2+6((3)/(2))-1

y=(7)/(2)

The vertex is:
$((3)/(2),\ (7)/(2))$

For a quadratic function the axis of symmetry is always a vertical line that passes through the vertex of the function.

Then the axis of symmetry is:

x=(3)/(2)