Practice!

Sample Problem

Find two quadratic functions (one that opens upwards and one that opens downward)

whose graphs have the x-intercepts: (-2, 0) and (3/2, 0).

f(x)=(x+1/4)2-(49/16)

f(x)=(x+1/4)2+(47/16)

f(x)=-(x+1/4)2-(47/16)

f(x)=(x-1/4)2+(1/8)

f(x)=-(x+1/4)2+(49/16)

Solution

With the knowledge that the parabola has the two x-intercepts (-2, 0) and (3/2, 0)

we can write an equation for a parabola that fulfills these conditions

(there are many parabolas that fulfill these conditions):

f(x)=(x+2)(x-3/2)=0

f(x)=x2+1/2x-3=0

f(x)=(x2+1/2x+1/16-1/16)-3=0 complete the square!

f(x)=(x2+1/2x+1/16)-1/16-3=0

f(x)=(x+1/4)2-(49/16)

What about a parabola that opens downward?

f(x)=(3/2-x)(x+2)=0 … a negative x2 term should make the parabola open downward

f(x)=-x2-1/2x+3=0

f(x)=-(x2+1/2x)+3

f(x)=-(x2+1/2x+1/16-1/16)+3

f(x)=-(x2+1/2x+1/16)-(-1/16)+3

f(x)=-(x+1/4)2+(49/16)