Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of each function. f(x)=6x^2+6x-5

Question

f(x)=6x^2+6x-5

Answer 1

VERTEX

For any parabola (quadratic) in this form f(x) = ax2 + bx + c (where a is not equal to 0), using the formula:

 

x = -b/2a

provides the x coordinate of the parabola's vertex.

So, for the quadratic in question, 6x² + 6x - 5,

x = -b/2a = -6/12 = -1/2

Then plugging this value of x in to the quadratic, we get

f(x) = 6(-1/2)² + 6(-1/2) - 5 = 6/4 - 6/2 - 5 = 3/2 - 3 - 5 = -1.5 - 5 = -6.5 (made the switch from fractions to a decimal)

The vertex of the parabola then is (-1/2, -13/2).

AXIS OF SYMMETRY

For a quadratic in standard form (like the one in question), the axis of symmetry is given by

x = -b/2a

Yes, just like the x coordinate of the vertex which makes sense since the axis of symmetry would have to pass through the vertex.

 

So, the axis of symmetry is x= -1/2.

Answer 2

Alright, Jas, remember that the formula for finding the axis of symmetry is -b/2a.

 

Plugging in what we have, that is -6/2(6) = -6/12 = -1/2

 

So, our axis of symmetry is x = -1/2

 

Now, we have to find the y-value if x = -1/2.

 

So, we get y = 6(-1/2)² + 6(-1/2) - 5 = 6(1/4) + 6(-1/2) - 5 = 6/4 - 6/2 - 5 = 3/2 - 3 - 5 = 3/2 - 6/2 - 10/2 = -13/2

 

This makes our vertex (-1/2, -13/2)

 

Hope this helps!