Question

How do you find the quadratic function with vertex (3,4) and point (1,2)?

Answer 1

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

Explanation:

A quadratic function can be written in vertex form as:

`f(x) = a(x-h)^2+k`

where `(h,k)` is the vertex and `a` is a constant multiplier.

In our example the vertex `(h, k)` is `(color(red)(3),color(blue)(4))`, so we can write:

`f(x) = a(x-color(red)(3))^2+color(blue)(4)`

Given that this passes through the point `(color(red)(1), color(blue)(2))`, we must have:

`color(blue)(2) = a(color(red)(1)-3)^2+4 = 4a+4`

Subtract `4` from both ends to get:

`-2 = 4a`

Divide both sides by `4` and transpose to find:

`a = -1/2`

So our quadratic function can be written in vertex form as:

`f(x) = -1/2(x-3)^2+4`

We can multiply this out and simplify as follows:

`f(x) = -1/2(x-3)^2+4`

`color(white)(f(x)) = -1/2(x^2-6x+9)+4`

`color(white)(f(x)) = -1/2x^2+3x-9/2+4`

`color(white)(f(x)) = -1/2x^2+3x-1/2`

Answer 2

`y=-1/2(x-3)^2+4`

Explanation:

The equation of a parabola in `color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`
where (h ,k) are the coordinates of the vertex and a is a constant.

here the vertex = (3 ,4) `rArrh=3,k=4`

Thus we can write `y=a(x-3)^2+4`

To find a, substitute the point (1 ,2) into the equation and solve for a

`x=1 : y=2rArr2=a(1-3)^2+4`

`rArr2=4a+4rArr4a=-2rArra=-1/2`

`"Thus " y=-1/2(x-3)^2+4larrcolor(blue)" in vertex form"`

distribute and simplify gives the equation in a different form.

`y=-1/2(x^2-6x+9)+4`

`rArry=-1/2x^2+3x-9/2+4`

`y=-1/2x^2+3x-1/2larrcolor(blue)" in standard form"`