Answer:
See Below.
Step-by-step explanation:
To convert a quadratic in standard form to vertex form, we will complete the square.
Let’s say we have a quadratic in standard form:
[tex]\displaystyle{f(x)=ax^2+bx+c[/tex]
To start, we will factor the leading coefficient from the first two terms:
[tex]\displaystyle {f(x)=a(x^2+\frac{b}{a}x)+c[/tex]
Next, we will divide the coefficient of the second term by 2, square it, and then add it to our equation.
Our coefficient of the second term is b/a. b/a divided by 2 is b/2a. And squaring yields b²/4a². So:
[tex]\displaystyle {f(x)=a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})+c[/tex]
Of couse, we will also need to subtract it as well to keep our equation equal.
Since a is being distributed, we will subtract a(b²/4a²) or b²/4a. So:
[tex]\displaystyle {f(x)=a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})+c-\frac{b^2}{4a}[/tex]
Finally, we can use the perfect trinomial pattern to factor. So:
[tex]\displaystyle {f(x)=a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}[/tex]
And this is vertex form.
Let’s see this with an example. Say we have:
[tex]f(x)=2x^2+4x-7[/tex]
And we want to convert this to vertex form.
As above, factor out the leading coefficient from the first two terms:
[tex]f(x)=2(x^2+2x)-7[/tex]
Divide the coefficent of the second term by 2 and then square it.
2/2 is 1. 1 squared is still 1.
So, we will add 1 within our parentheses:
[tex]f(x)=2(x^2+2x+1)-7[/tex]
Since we added 1 inside, we must subtract 2(1) outside. So:
[tex]f(x)=2(x^2+2x+1)-7-2[/tex]
Now, we can factor. Therefore:
[tex]f(x)=2(x+1)^2-9[/tex]
And this is in vertex form.
Answer:
A quadratic function in standard form is converted to vertex form by completing the square. The first two terms are used to create a perfect square trinomial after a zero pair is added. The zero pair is found by taking half of the x-term coefficient and squaring it. The original constant term and the negative value of the zero pair are then combined.
Step-by-step explanation:
its the sample response
