Question

How do I graph quadratics in vertex form?

Answer 1

Explanation:

Equation of the quadratic function given as,

g(x) =
`(1)/(3)(x-6)^2+1`

Since, leading coefficient of the function is positive,

Parabola will open upwards.

For x- intercepts,

g(x) = 0

`(1)/(3)(x-6)^2+1=0`

(x - 6)² = -3

(x - 6) = ±√(-3)

x = 6 ± √(-3)

Since, x is an imaginary number, this function has no x-intercepts.

Similarly, for y-intercept,

x = 0

g(0) =
`(1)/(3)(0-6)^2+1`

= 12 + 1

= 13

y-intercept of the function → y = 13

Vertex of the parabola → (6, 1)

For x = 12,

g(12) =
`(1)/(3)(12-6)^2+1`

= 13

Therefore, one point lying on the graph will be (12, 13).

Now we can graph the quadratic function given in the question.