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

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Answer 1 · 2018-05-09T21:23:50

show the explanation below

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]}