Question

Write the quadratic function f(x) = x2 + 8x + 3 in vertex form? A) f(x) = (x - 4)2 - 13 B) f(x) = (x - 4)2 + 3 C) f(x) = (x + 4)2 + 3 D) f(x) = (x + 4)2 - 13

Answer 1

`"D": f(x)=(x+4)^2-13`

Explanation:

Given the following function, you are asked to convert it to vertex form:

`f(x)=x^2+8x+3`

The given possible solutions are:

`"A") f(x)=(x-4)^2-13`
`"B") f(x)=(x-4)^2+3`
`"C") f(x)=(x+4)^2+3`
`"D") f(x)=(x+4)^2-13`

Converting to Vertex Form
`1`. Start by placing brackets around the first two terms.

`f(x)=x^2+8x+3`

`f(x)=(x^2+8x)+3`

`2`. In order to make the bracketed terms a perfect square trinomial, we must add a "`color(darkorange)c`" term as in `ax^2+bx+color(darkorange)c`. Since `color(darkorange)c`, in a perfect square trinomial is denoted by the formula `color(darkorange)c=(color(blue)b/2)^2`, take the value of `color(blue)b` to find the value of `color(darkorange)c`.

`f(x)=(x^2+color(blue)8x+(color(blue)8/2)^2)+3`

`3`. However, adding `(8/2)^2` would change the value of the equation. Thus, subtract `(8/2)^2` from the `(8/2)^2` you just added.

`f(x)=(x^2+8x+(8/2)^2-(8/2)^2)+3`

`4`. Multiply `(-(8/2)^2)` by the `color(violet)a` term as in `color(violet)ax^2+bx+c` to bring it outside the brackets.

`f(x)=(color(violet)1x^2+8x+(8/2)^2)+3-((8/2)^2xxcolor(violet)1)`

`5`. Simplify.

`f(x)=(x^2+8x+16)+3-16`

`f(x)=(x^2+8x+16)-13`

`6`. Lastly, factor the perfect square trinomial.

`color(green)(|bar(ul(color(white)(a/a)f(x)=(x+4)^2-13color(white)(a/a)|)))`