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How do I graph quadratics in vertex form?

How Do I Graph Quadratics In Vertex Form class=

Sagot :

Answer:

Step-by-step explanation:

Equation of the quadratic function given as,

g(x) = [tex]\frac{1}{3}(x-6)^2+1[/tex]

Since, leading coefficient of the function is positive,

Parabola will open upwards.

For x- intercepts,

g(x) = 0

[tex]\frac{1}{3}(x-6)^2+1=0[/tex]

(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) = [tex]\frac{1}{3}(0-6)^2+1[/tex]

      = 12 + 1

      = 13

y-intercept of the function → y = 13

Vertex of the parabola → (6, 1)

For x = 12,

g(12) = [tex]\frac{1}{3}(12-6)^2+1[/tex]

        = 13

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

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

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