Question

How do you graph the parabola `y= 3(x+4)^2+5` using vertex, intercepts and additional points?

Answer 1

show the explanation below

Explanation:

firstly we will simplified it

`y= 3(x+4)^2+5`

`y=3(x^2+8x+16)++5`

`y=3x^2+24x+48+5`

`y=3x^2+24x+53`

we will compare it with

`y=ax^2+bx+c`

we will get

`a=3,b=24and c=53`

now we will find the vertix

`x_v=-b/(2a)`

`x_v=-24/(2*3)=-24/6=-4`

`y_v=3(-4)^2+24(-4)+53=48-96+53=101-96=5`

vertix `(-4,5)`

to find the intercept

suppose `x=0`

`y=0+0+53=53`

the y intercept `(53,0)`

use some points

`f(2)=12+48+53=113`

`f(1)=80`

`f(0)=53`

`f(_1)=32`

`f(-2)=17`

now we will skitch the function `y=3x^2+24x+53`

graph{3x^2+24x+53 [-11.22, 11.28, 1.51, 12.76]}