# Standard Form

### Jananie Chandranathan

## Learning Goals!

Being able to go from standard form to vertex form by completing the square

Being able to use standard form to determine the y-intercept

## Summary

In this unit so far, we have learned 2 different ways to represent a parabola. Vertex form and factored form. Today, I will be showing you the final way to represent a parabola. This way is called the standard form. Unlike the vertex form and the factored form, the standard form can only show what the y-intercept is. The equation for the standard for is y=ax^2+bx+c , in which c is the y-intercept. Again the value of a represents the direction of opening of the parabola, just like all the other a's in vertex form and factored form. If a is positive, the graph opens upward, if a is negative, the graph opens downward. By plugging in zero into the x values of the equation, c will be left over and it will represent the y intercept.

## How to do it?

The purpose to go from standard form to vertex form is so that we can determine the vertex of the parabola. With just the standard form, all we can do is determine the y-intercept. To find the vertex form with the standard form equation, we must do a little trick called "completing the square"

## Example

Lets say we have an equation of a parabola which is (y=x^2+12x+32)

For us to convert this standard form to vertex form, we are going to complete the square.

Here are the Steps!

1. Move the c value of the equation to the other side: -32+y = x^2 +12x

2. Now if you look at your right side, the b value is a 12. We must now add a number next to bx so that when we factor the right side, we will get a square. So lets add 36 since factoring the right side will give us (x+6)^2. Also if u add 36 to once side u must add it to the other. : -32+y+36 = x^2 + 12x + 36

3. Now we simplify! 4+y = (x+6)^2.

4. Now we bring 4 to the right side to isolate y : y = (x+6)^2 - 4

5. Now we have the vertex form!

## Quadratic Formula

The quadratic formula is a very useful formula for determining the zeros of a parabola. The equation is :

## How to use it?

Using this formula is very simple. All you have to do is sub in the values for a, b, c and you will get 2 values which will be your 2 zeros.

## Word Problem (Completing the Square)

Problem: The path of a ball is modelled by the equation y = −x+2x+ 3, where x is the horizontal distance, in metres, from a fence and y is the height, in metres, above the ground.

a) What is the maximum height of the ball?

y=-x^2+2x+3

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

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

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

y=-(x-1)^2+4

The maximum height of the ball is 1m.

b) At what horizontal distance does the ball reach its maximum height?

It reaches its maximum height at 4m.