Standard Form--> Vertex Form
x^2+4x+4
(x+2)^2
(x+3)^2
x^2+6x+9
Solve
0=x^2+4x-5
x=-5
or
x=1
How many real solutions does f(x)=x^2-8x+3 have?
2 real solutions
(-8)^2-4(1)(3)
64-12
52>0
Minimum at (1,5)
f(x)=(x-1)^2+5
x^2+4x+5
(x+2)^2+1
(x+6)^2-4
x^2+12x+32
Solve
0=(x+4)^2-25
x=1
or
x=-9
How many real solutions does h(x)=2x^2-7x+10 have?
0 real solutions
(-7)^2-4(2)(10)
49-80
-31<0
Minimum at (4,-7)
f(x)=(x-4)^2-7
x^2+16x-7
(x+8)^2-71
(x-5)^2+7
x^2-10x+32
Solve
0=(x-7)^2+81
x=7+-9i
g(x)=2x^2-10x+c
Find a value for c that allows g(x) to have 2 real solutions.
Any number that is less than 12.5
c<12.5
Maximum at (-6,10)
f(x)=-(x+6)^2+10
x^2-14x-3
(x-7)^2-52
(x-11)^2-8
x^2-22x+113
Solve
0=x^2-12x+27
x=6+-3i
g(x)=2x^2-10x+c
Find a value for c that allows g(x) to have 1 real solutions.
c=12.5
Minimum at (-13,-2)
f(x)=(x+13)^2-2
x^2-5x+12
(x-2.5)^2+5.75
(x-5/2)^2+23/4
3(x+2)^2+9
3x^2+12x+21
Solve
-212=3(x+5)^2+220
x=-5+-12i
g(x)=2x^2-10x+c
Find a value for c that allows g(x) to have 0 real solutions.
Any number that greater than 12.5
c>12.5
Maximum at (-1/2,4)
f(x)=-(x+1/2)^2+4