# Standard Form

## Learning goal

1. y= x^2+2x+1, Complete the square

2. y= 4x^2+8x+16, Use quadratic formula to find x- intercept

3. using the equation, 4x^2+8x+16, graph the parabola

## Summary

Standard form is y=ax^2+bx+c, where the a,b and c are known values, a cant be 0. X is the variable, that we dont know yet. the a value gives you the general shape and the direction which the parabola opens. the c value is the y-intercept. you can use the quadratic formula to find the x-intercepts or complete the square to find the maximum or minimum value

Quadratic Formula is a formula where you sub in the a,b and c values so that you can find the x-intercepts of the parabola. you can also use the discriminate to see, how many x-intercepts a parabola has. If the D>0, it will have 2 x-intercepts, if D<0, it will have 0 x-intercepts, and if the D=0 there will be only one x- intercept

y= 2x^ + 9x+ 6

x= -9+-√ 9^2-4(2)(6)/ 2(2)

x= -9+-√ 81-48/4

x= -9+-√ 33/4

x= -9+-5.7/4

x=-9+5.7/4=-0.825

x= -9-5.7/4= 3.675

## Completing the Square

y= x^2 +6x+2

y=( x^2+ 6x+9-9) +2

y= (x+3)^2 -9+2

y=(x+3)^2-7

## Word Problem

h= -0.25d^2+2d+1.5

1.h is the height and d is the distance in meters, what is the maximum height of the ball and at what horizontal distance does this occur

h=-0.25(d^2- 8d+16-16) +1.5

h= -0.25(d-4)^2 +4+1.5

h= -0.25(d-4)^2 +5.5

therefore the max height is 5.5 m and it occurs at 4 m