Question

What is the standard form of the quadratic function that has a vertex at (3,4)and goes through the point (4,5)?A. y= 2x2 - 12x+ 22O B. y = x - 6x + 13O c. y = y2 + 6x + 5O D. y= x2 - 6x + 9SUBMIT

Answer 1

The question asks us to find the standard form of the quadratic function that has a vertex at (3, 4) and goes through (4, 5).

In order to solve this question, we simply use the general vertex form of a quadratic equation.

The vertex form a quadratic equation is:

`\begin{gathered} y=a(x-h)^2+k \\ \text{where,} \\ (h,k)\text{ is the vertex of the quadratic equation} \end{gathered}`

We have been given the vertex coordinates as (3, 4).

`\therefore h=3,k=4`

Thus, the incomplete quadratic equation:

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

Now, we need to find the value of "a".

In order to find "a", we can use the coordinates that the quadratic graph passes through

i.e. (4, 5)

This is done below:

`\begin{gathered} y=a(x-3)^2+4 \\ at\text{ point (4, 5),} \\ x=4,y=5 \\ \\ \therefore5=a(4-3)^2+4 \\ 5=a+4 \\ \text{subtract 4 from both sides} \\ 5-4=a+4-4 \\ \therefore a=1 \end{gathered}`

Now that we have the value of "a", we can write the quadratic equation as:

`\begin{gathered} y=a(x-h)^2+k \\ y=1*(x-3)^2+4 \\ y=(x-3)^2+4 \\ \text{Expand the bracket} \\ y=(x^2-3x-3x+9)+4 \\ y=x^2-6x+9+4 \\ \therefore y=x^2-6x+13\text{ (Option B)} \end{gathered}`

Therefore, the final answer is:

`y=x^2-6x+13\text{ (Option B)}`