Question

What is the vertex and the equation of the axis of symmetry of the graph of y=x^2-6x-7

A. vertex: (–16, 3)
a.o.s: x = –16

B. vertex: (–3, 20)
a.o.s: x = –3

C. vertex: (3,–16)
a.o.s: x = 3

D. vertex: (20, –3)
a.o.s: x = 20

Answer 1

Using the formula x = -b/(2a) We get x = 6/2=3 There is only one choice with x =3 so c is the answer

Answer 2

Option C is correct

Vertex: (3, -16) and axis of symmetry is x = 3

Explanation:

The axis of symmetry for a parabolic general equation
`y=ax^2+bx+c` is given by:

`x = -(b)/(2a)` ....[1]

Vertex of the equation is:
`(-(b)/(2a), f(-(b)/(2a))`

As per the statement:

The graph of the equation is:

`y=f(x) = x^2-6x-7` ....[2]

On comparing with general parabolic equation i.e
`y=ax^2+bx+c` we have;

a = 1, b = -6 and c = -7

By [1] we have;

`x = -(b)/(2a)`

Substitute the given values we have;

`x = -(-6)/(2(1)) =(6)/(2)`

Simplify:

x = 3

Substitute the value of x= 3 in [2] we have;

`f(3) = 3^2-6(3)-7=9-18-7=-9-7=-16`

Therefore, the vertex of the given equation is: (3, -16) and axis of symmetry is x = 3