Question

Explain the steps necessary to convert a quadratic function in standard form to vertex form

Answer 1

Answer:

See below.

Step-by-step explanation:

Here's an example to illustrate the method:

f(x) = 3x^2 - 6x + 10

First divide the first 2 terms by the coefficient of x^2 , which is 3:

= 3(x^2 - 2x) + 10

Now  divide the -2 ( in -2x) by 2 and write the x^2 - 2x in the form

(x - b/2)^2 - b/2)^2  (where b = 2) , which will be equal to x^2 - 2x in a different form.

= 3[ (x - 1)^2 - 1^2 ] + 10 (Note: we have to subtract the 1^2 because (x - 1)^2 = x^2 - 2x  + 1^2  and we have to make it equal to x^2 - 2x)

= 3 [(x - 1)^2 -1 ] + 10

= 3(x - 1)^2 - 3 + 10

= 3(x - 1)^2 + 7 <------- Vertex form.

In general form the vertex form of:

ax^2 + bx + c  = a [(x - b/2a)^2 - (b/2a)^2] + c .

This is not easy to commit to memory so I suggest the best way to do these conversions is to remember the general method.

Answer 2

Answer :

Vertex form

$a\left\{(x+\frac{b}{2a})^2-(\frac{b}{2a})^2\right\}+c$

Step-by-step explanation:

We are given than a quadratic  function in standard form

$ax^2+bx+c$

We have to explain steps which is necessary for converting quadratic x function in standard form to vertex form

We are explaining steps for converting a quadratic function into vertex form with the help of example

Suppose we have a quadratic function

$2x^2+x-1$

Taking 2 common from the given function the we get

$2(x^2+\frac{x}{2})-1$

Now, we convert the equation of the form $(a+b)^2 or (a-b)^2$

$2\left \{(x)^2+2\times x\times\frac{1}{4}+\frac{1}{16}-\frac{1}{16}\right\}-1$

$2\left\{(x+\frac{1}{4})^2-\frac{1}{16}\right\}-1$

$2\left\{(x+\frac{1}{4})^2-(\frac{1}{4})^2\right\}-1$

Vertex form=$2\left\{(x+\frac{1}{4})^2-(\frac{1}{4})^2\right\}-1$

Hence , the vertex form=$a\left\{(x+\frac{b}{2a})^2-(\frac{b}{2a})^2\right\}+c$