Answer:
a. a = 1, b = -5, c = -14
b. a = 1, b = -6, c = 9
c. a = -1, b = -1, c = -3
d. a = 1, b = 0, c = -1
e. a = 1, b = 0, c = -3
Step-by-step explanation:
a. x-ints at 7 and -2
this means that our quadratic equation must factor to:
[tex](x-7)(x+2) = 0[/tex]
FOIL:
[tex]x^2 + 2x - 7x - 14 = 0[/tex]
Simplify:
[tex]x^2 - 5x - 14 = 0[/tex]
a = 1, b = -5, c = -14
b. one x-int at 3
this means that the equation will factor to:
[tex](x-3)^2=0[/tex]
FOIL:
[tex]x^2 - 3x - 3x +9 = 0[/tex]
Simplify:
[tex]x^2 - 6x + 9 = 0[/tex]
a = 1, b = -6, c = 9
c. no x-int and negative y must be less than 0
This means that our vertex must be below the x-axis and our parabola must point down
There are many equations for this, but one could be:
[tex]-x^2-x-3=0[/tex]
a = -1, b = -1, c = -3
d. one positive x-int, one negative x-int
We can use any x-intercepts, so let's just use -1 and 1
The equation will factor to:
[tex](x+1)(x-1)=0[/tex]
This is a perfect square
FOIL:
[tex]x^2-1=0[/tex]
a = 1, b = 0, c = -1
e. x-int at [tex]\sqrt{3} , -\sqrt{3}[/tex]
our equation will factor to:
[tex](x+\sqrt{3} )(x-\sqrt{3)} =0[/tex]
This is also a perfect square
FOIL and you will get:
[tex]x^2 - 3 = 0[/tex]
a = 1, b = 0, c = -3
