Ask your own question, for FREE!
Mathematics 6 Online
OpenStudy (anonymous):

An expression is shown below: f(x) = -16x2 + 24x + 16 Part A: What are the x-intercepts of the graph of f(x)? Show your work. Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph.

OpenStudy (cggurumanjunath):

b- maximum

OpenStudy (anonymous):

why ?

OpenStudy (cggurumanjunath):

x intercept =-b/2a =-24/(2*-16)=24/32=12/16=6/8=3/4=0.75

OpenStudy (cggurumanjunath):

b- maximum because " a " =-16 which is < (less) than zero !

OpenStudy (anonymous):

whoaaa lol u sure

OpenStudy (cggurumanjunath):

x=0.75 ;find y by inputting x in f(x)

OpenStudy (anonymous):

it never mentioned an "a" tho

OpenStudy (cggurumanjunath):

-16(0.75)^2+24*(0.75)+16

OpenStudy (cggurumanjunath):

f(x)=-16x^2+24x+16 f(x)=ax^2+bx+c

OpenStudy (anonymous):

where did the a come from

OpenStudy (cggurumanjunath):

a=-16 b=24 c=16

OpenStudy (cggurumanjunath):

f(x)=ax^2+bx+c - quadratic equation

OpenStudy (cggurumanjunath):

on comparing we get "a" ! is it clear ?

OpenStudy (anonymous):

if u say so

OpenStudy (anonymous):

and part c ?

OpenStudy (cggurumanjunath):

coordinates of vertex are (x,y)

OpenStudy (cggurumanjunath):

f(x)=25

OpenStudy (cggurumanjunath):

vertex is (0.75,25)

OpenStudy (cggurumanjunath):

u know how to plot the graph right ?

OpenStudy (anonymous):

no

OpenStudy (cggurumanjunath):

find f(x ) by assuming x=0.5,x=1 corresponging values of f(x) can be found similarly

OpenStudy (cggurumanjunath):

and on x and y axis of coordinate system plot those (x,y) points and draw a smooth curve which is your graph !

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!