Question

What is the standard form of the quadratic function that has a vertex at (4, 5) and goes through the point (5, 6)?

Answer 1

the standard form of the quadratic function that meets the given conditions is:

`\[ y = x^2 - 8x + 21 \]`

Therefore, the answer is option B.

To find the standard form of the quadratic function
`\(y = ax^2 + bx + c\)` that has a vertex at
`\((h, k) = (4, 5)\)` and passes through the point
`\((x, y) = (5, 6)\)`, we can use the vertex form of a quadratic function and then convert it to standard form. The vertex form of a quadratic function is:

`\[ y = a(x - h)^2 + k \]`

Given the vertex
`\((h, k) = (4, 5)\)`, we have:

`\[ y = a(x - 4)^2 + 5 \]`

Now we need to determine the value of \(a\) using the point \((5, 6)\):

`\[ 6 = a(5 - 4)^2 + 5 \]`

`\[ 6 = a(1)^2 + 5 \]`

`\[ a = 6 - 5 \]`

`\[ a = 1 \]`

So the equation in vertex form is:

`\[ y = (x - 4)^2 + 5 \]`

To convert this to standard form, we expand the squared term:

`\[ y = (x^2 - 8x + 16) + 5 \]`

`\[ y = x^2 - 8x + 21 \]`

Thus, the standard form of the quadratic function that meets the given conditions is:

`\[ y = x^2 - 8x + 21 \]`

Therefore, the answer is option B.

Answer 2

The vertex form of the equation of a quadratic function is given to be:

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

where (h, k) is the vertex.

From the question, we have the following parameters:

`\begin{gathered} (h,k)=(4,5) \\ (x,y)=(5,6) \end{gathered}`

Therefore, the equation will be in the form:

`y=a(x-4)^2+5`

At the point (5, 6), we can get the value of a to be:

`\begin{gathered} 6=a(5-4)^2+5 \\ a=1 \end{gathered}`

Therefore, we have the vertex form to be:

`y=(x-4)^2+5`

Expanding, we have the equation to be:

`\begin{gathered} y=x^2-8x+16+5 \\ y=x^2-8x+21 \end{gathered}`

OPTION B is correct.