# Quadratic (Parabolic) Function - Convert from Standard to Vertex Form - A not 1

## Quadratic Function Convert from Standard to Vertex Form - How it Works - Video

### Definitions

Definitions:

The first equation that we see for the quadratic function is f(x) = ax2 + bx + c.

We can transform that standard equation into another one called the vertex form, f(x) = a(x - h)2 + k. We can use this form to graph our equation better.

We have two options for a parabola: opening upward or opening downward.

### Quadratic Model

Quadratic Model:

Here we have two ways for our parabolic function to open where the symmetry is along the vertical axis.

On the left the parabolic function open upwards, because a > 0.

On the right the parabolic function open downwards, because a < 0.

### Theorems

Theorems:

The first theorem that is useful is how to find the vertex if our equation is in standard form. The x-coordinate or h can be found be using -b/ (2a) and the y-coordinate or k can found by substituting that value into the function.

The second theorem that is useful is how to find the maximum or minimum. The result after substituting will be our maximum or our minimum depending if a is positive or negative. If a < 0 then we have a maximum. If we have a > 0, then we have a minimum.

### Quadratic Guide

Quadratic Guide:

Here we have a chart on how to find vertex and the x-intercepts.

### Example 1

Example 1:

Here we have a standard parabolic equation, f(x) = x2 - 3x + 5.

Line 1 - Our equation.

Line 2 - We added (b/a)2 to both sides of the equation.

Line 3 - We substituted -3 for b.

Line 4 - We squared our term.

Line 5 - We subtracted 9/4 on both sides.

Line 6 - We factored and subtracted the constants.

Now we have our equation in vertex form, f(x) = (x - 3/2)2 + 11/4.

Now let's graph.

We have vertex form, f(x) = (x - 3/2)2 + 11/4, which means we have one point - the vertex, (3/2, 11/4). There are multiply ways to find other points: X/Y Table, finding the x-intercepts, find the y-intercepts, or just substituting a few numbers.

Here we are going to find the x-intercepts in order to do that we need to substitute 0 for f(x).

f(x) = (x - 3/2)2 + 11/4

0 = (x - 3/2)2 + 11/4

-11/4 = (x - 3/2)2

±sqrt(-11/4) = x - 3/2

Ø

Substituted 0 for f(x).

Subtracted -11/4 on both sides.

Square rooted both sides.

No solution is radicand is a not real result.

Since we have a negative underneath square root symbol, our graph does not cross the x-axis. Ah, that means we need to choose another way to graph our equation.

Since we don't have x-intercepts, we have to do another way to find points. Let's do an X/Y Table.

As we have our equation in vertex form, f(x) = (x - 3/2)2 + 11/4, we have one point already, (3/2, 11/4). Since parabolic equations are symmetric, we want to pick one point above the extrema and one point below the extrema. Make sure that they are equal distance from the vertex, because they should be same result.

We picked 1/2 and 5/2 for out inputs and our result for both is 15/4 because

f(x) = (x - 3/2)2 + 11/4

f(1/2) = (1/2 - 3/2)2 + 11/4

= (2/2)2 + 11/4

= (1)2 + 11/4

= 1 + 11/4

= 4/4 + 11/4

= 15/4

Given

Substituted 1/2 for x.

Subtracted.

Divided.

Squared.

Found common denominator.

Added.

f(x) = (x - 3/2)2 + 11/4

f(1/2) = (5/2 - 3/2)2 + 11/4

= (2/2)2 + 11/4

= (1)2 + 11/4

= 1 + 11/4

= 4/4 + 11/4

= 15/4

Given

Substituted 5/2 for x.

Subtracted.

Divided.

Squared.

Found common denominator.

Added.

Now we have two more points. Since we used the X/Y Table. The vertex is our third point. Since a > 0, we have a minimum. Now we just connect the dots.

### Example 2

Example 2:

Here we have a standard parabolic equation, f(x) = -x2 - 4x + 2.

Right now we have a problem. We need a to be +1 to complete the square in order to for that to happen we need to factor out that negative.

Line 1 - Our equation.

Line 2 - On the right side we factored the negative out so a would be +1. We added (b/a)2 to both sides of the equation but on the left we multiplied (b/a)2 by the term outside the parenthesis on the right, -1. We have to do this because we need symmetry on both sides. If we redistributed the negative on the right side the term, (b/a)2, will be negative. That is why we have to multiply what we added on the left side by -1 so we have added the same term to both sides.

Line 3 - We substituted 4 for b. Notice how b is not negative. We used the coefficients inside the parenthesis.

Line 4 - We squared our term.

Line 5 - We added 4 on both sides.

Line 6 - We factored and added the constants.

Now we have our equation in vertex form, f(x) = -(x + 2)2 + 6.

Now let's graph.

We have vertex form, f(x) = -(x + 2)2 + 6, which means we have one point - the vertex, (-2, 6). There are multiply ways to find other points: X/Y Table, finding the x-intercepts, find the y-intercepts, or just substituting a few numbers.

Here we are going to find the x-intercepts in order to do that we need to substitute 0 for f(x).

f(x) = -(x + 2)2 + 6

0 = -(x + 2)2 + 6

-6 = -(x + 2)2

6 = (x + 2)2

±sqrt(6) = x - 2

x = -2 ± sqrt(6)

Given

Substituted 0 for f(x).

Subtracted 6 on both sides.

Divided by -1 on both sides.

Square rooted both sides.

Added 1 to both sides

Now we have two more points. Since we substituted 0 for f(x)/y, we just found our x-intercepts. So one point is (-2 - sqrt(6), 0) and our point is (-2 + sqrt(6), 0)

The vertex is our third point. Since a < 0, we have a maximum. Now we just connect the dots.

### Example 3

Example 3:

Here we have a standard parabolic equation, f(x) = 3x2 - 6x - 3.

Right now we have a problem. We need a to be +1 to complete the square in order to for that to happen we need to factor out that +3.

Line 1 - Our equation.

Line 2 - On the right side we factored the +3 out so a would be +1. We added (b/a)2 to both sides of the equation but on the left we multiplied (b/a)2 by the term outside the parenthesis on the right, +3. We have to do this because we need symmetry on both sides. If we redistributed the +3 on the right side the term, (b/a)2, will be +3 * (b/a)2. That is why we have to multiply what we added on the left side by +3 so we have added the same term to both sides.

Line 3 - We substituted -2 for b. This time the b is the same, and this is because we did not have to factor out a negative number.

Line 4 - We squared our term.

Line 5 - We subtracted 3 on both sides.

Line 6 - We factored and added the constants.

Now we have our equation in vertex form, f(x) = 3(x - 1)2 - 6.

Now let's graph.

We have vertex form, f(x) = 3(x - 1)2 - 6, which means we have one point - the vertex, (1, 6). There are multiply ways to find other points: X/Y Table, finding the x-intercepts, find the y-intercepts, or just substituting a few numbers.

Here we are going to find the x-intercepts in order to do that we need to substitute 0 for f(x).

f(x) = 3(x - 1)2 - 6

0 = 3(x - 1)2 - 6

6 = 3(x - 1)2

2 = (x - 1)2

±sqrt(2) = x - 1

x = 1 ± sqrt(2)

Given

Substituted 0 for f(x).

Added 6 on both sides.

Divided by 3 on both sides.

Square rooted both sides.

Added 1 to both sides

Now we have two more points. Since we substituted 0 for f(x)/y, we just found our x-intercepts. So one point is (1 - sqrt(2), 0) and our point is (1 + sqrt(2), 0)

The vertex is our third point. Since a > 0, we have a minimum. Now we just connect the dots.