vertex form help

Question

Write a quadratic function in the vertex form f(x)=a(x-h)² +k such that the graph of the function opens down and has a vertex at (10,-2) and passes through point (9,-3)



Answer 1

You already wrote down the vertex form for a quadratic function f(x)=a(x-h)²+k. Let's review what the values mean in the vertex form.

a - how vertically stretched/compressed the parabola is. a positive value --> parabola faces up. a negative value --> parabola faces down (flipped)

h - x value of the vertex

k = y value of the vertex

You are given the vertex of the parabola as (10,-2).

Plug these values into h, k in the vertex form.

This gives you f(x)=a(x-10)²-2.

Now you need to figure out the value of a.

How to do this?

Remember you were given another point the parabola passes through: (9,-3).

This means that

in f(x)=a(x-10)²-2,

f(9) = a(9-10)²-2 = -3.

Now all you need to do is solve for the value of a in the equation above.

Once you do that, fill in the value of a back into the original equation f(x)=a(x-10)²-2.

Then you have a complete answer.

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Answer 2

vertex = max = (10,-2) through point (9,-3)

y = a(x-h)^2 + k with (h,k) = (10,-2)

y= a(x-10)^2 -2, plug in (9,-3) to solve for "a"

-3 = a(9-10)^2-2, a=-1

y = -(x-10)^2 -2 in vertex form

Answer 3

a(x-h)^2 + k is the vertex form.

(h,k) is the vertex. In this case, that's (10,-2)

y = a(x-10)^2 - 2

Plug the point in to find the value of a.

(9, -3)

-3 = a(9-10)^2 -2

-1 = a(-1)^2

-1 = 1a

-1=a

The function is y = -(x-10)^2 - 2