Explain who to write a quadratic equation that has -6 as its only root
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Answer:
see below
Step-by-step explanation:
f(x) = −16x^2 + 22x + 3
Factor out the negative
f(x) = -( 16x^2 -22x -3)
= -(8x+1)(2x-3)
Find the x intercepts
Set y = 0
0 = -(8x+1)(2x-3)
Using the zero product property
8x+1 =0 2x-3 = 0
8x = -1 2x = 3
x = -1/8 x =3/2
The x intercepts are ( -1/8, 0) and ( 3/2, 0)
The end behavior
-16 x^2 is the dominate term
Let x →-∞
f(-∞) = -16 (-∞)^2 = -16 (∞) = -∞
As x goes to negative infinity y goes to - infinity
Let x →∞
f(∞) = -16 (∞)^2 = -16 (∞) = -∞
As x goes to infinity y goes to - infinity
We know this is a downward facing parabola a < 0 and this is a quadratic
We have the x intercepts
We can find the axis of symmetry from the zeros
(-1/8+ 3/2) /2 = (-1/8 + 12/8)/2 = (11/8)/2 = 11/6
The axis of symmetry is x = 11/16
Using the axis of symmetry and the equation, we can find the maximum point
y = -(8*11/16+1)(2*11/16-3) = 169/16
The vertex is at (11/16, 169/16(
