Standard Form

Quadratic Formula, and Completing the square

Learning Goals

Here are some of the learning goals for when using standard form.


1) You will learn how to find the x intercepts of a parabola using the quadratic formula.

2) You will learn how to complete the square in order to find the vertex of a parabola.

Summary

What is the quadratic formula?

The quadratic formula is

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The quadratic formula is used to find the x intercepts of a parabola.

The values for the variables a,b, and c are taken from standard form quadratic equation.

The standard form quadratic equation is


y=ax^2+bx+c


Inside the quadratic formula, there is something called the discriminant.

The discriminant is

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If D=0, there is only 1 x-intercept.

If D>0, there are 2 x-intercepts.

If D<0, there are no x-intercepts.


What is Completing the square?


Completing the square is the method in which the standard form quadratic equation is turned into the vertex form quadratic equation to find the vertex of the parabola.


ax^2+bx+c ----> a(x-h)^2+k

Example of Quadratic Formula

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The first thing that was done was the a, b, and c values were taken from the standard form equation.

They were then substituted into the respective variables in the quadratic formula.

Now to solve.


You can see that 2 x-intercepts were found. You can see this because the discriminant was a greater than 0.

Example 1: Using the quadratic formula | Quadratic equations | Algebra I | Khan Academy

Example of Completing the Square

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By using the completing the square method you can see that the standard form equation

y=2x^2+16x-3 was turned into the vertex form quadratic equation y=2(x+4)^2-35.


This allows us to find the vertex of this parabola, which is (-4,-35).

Word Problem (Completing the Square)

A ball was thrown from a ladder. The equation that the ball makes when thrown is y=-5x^2+10x+1. The height is represented by h in metres, and time is represented by x in seconds. What is the maximum height the ball reached, and what time did the ball reach this height?


y=-5x^2+10x+1 (-2/2)^2 = -1^2 = 1

y=(-5t^2+10x)+1

y=-5(x^2-2x)+1

y=-5(x^2-2x+1-1)+1

y=-5(x^2-2x+1)+1+5

y=-5(x^2-2x+1)+6

y=-5(x-1)^2+6


h=1, k=6


Therefore the ball reached the maximum height of 6 metres after 1 second.