Question

algebra 2
3.3 - day 3 - convert quadratics from standard form to vertex form
use the information provided to write the vertex - form equation of each parabola. identify the vertex.

1. (f(x)=x^{2}-10x + 25)
2. (f(x)=x^{2}+8x + 11)
3. (f(x)=x^{2}+16x + 57)
4. (f(x)=x^{2}-16x + 63)

Explanation:

Step1: Recall vertex - form of quadratic

The vertex - form of a quadratic function is $f(x)=a(x - h)^2+k$, where $(h,k)$ is the vertex of the parabola. For a quadratic function in standard form $f(x)=ax^{2}+bx + c$, we can complete the square to convert it to vertex - form.

Step2: Convert $f(x)=x^{2}-10x + 25$ to vertex - form

We know that $x^{2}-10x + 25=(x - 5)^{2}+0$. Here $a = 1$, $h = 5$, and $k = 0$. So the vertex - form is $f(x)=(x - 5)^{2}$ and the vertex is $(5,0)$.

Step3: Convert $f(x)=x^{2}+8x + 11$ to vertex - form

First, complete the square for $x^{2}+8x$. We know that $(x + m)^{2}=x^{2}+2mx+m^{2}$. For $x^{2}+8x$, we have $2m = 8$ (so $m = 4$) and $x^{2}+8x=(x + 4)^{2}-16$. Then $f(x)=x^{2}+8x + 11=(x + 4)^{2}-16+11=(x + 4)^{2}-5$. The vertex is $(-4,-5)$.

Step4: Convert $f(x)=x^{2}+16x + 57$ to vertex - form

Complete the square for $x^{2}+16x$. Since $2m=16$ (so $m = 8$), $x^{2}+16x=(x + 8)^{2}-64$. Then $f(x)=x^{2}+16x + 57=(x + 8)^{2}-64+57=(x + 8)^{2}-7$. The vertex is $(-8,-7)$.

Step5: Convert $f(x)=x^{2}-16x + 63$ to vertex - form

Complete the square for $x^{2}-16x$. Since $2m=-16$ (so $m=-8$), $x^{2}-16x=(x - 8)^{2}-64$. Then $f(x)=x^{2}-16x + 63=(x - 8)^{2}-64+63=(x - 8)^{2}-1$. The vertex is $(8,-1)$.

Answer:

1. Vertex - form: $f(x)=(x - 5)^{2}$, Vertex: $(5,0)$
2. Vertex - form: $f(x)=(x + 4)^{2}-5$, Vertex: $(-4,-5)$
3. Vertex - form: $f(x)=(x + 8)^{2}-7$, Vertex: $(-8,-7)$
4. Vertex - form: $f(x)=(x - 8)^{2}-1$, Vertex: $(8,-1)$