Identify the intervals on each quadratic function is positive

Question

y=x^2+2x-8

Y=x^2-5x-24

Y=-x^2+4x+12

Y=5x^2-3x-8

Answer 1

−For a quadratic function ax² + bx + c where coefficient a is positive (which is the case for each of the four given functions), then the function will be negative in between the zeroes, if they exist, zero at the zeroes (of course!), and positive everywhere else. You can find zeroes by factoring or by using the quadratic formula.

y = x² + 2x – 8 = (x + 4)(x – 2) so y = 0 when x = −4 or x = 2 as Denise G. showed in her earlier response.

So y < 0 on the interval (−4,2) and positive on the intervals (−∞,–4) and (2,∞)

y = x² − 5x – 24 = (x + 3)(x – 8) so y = 0 when x = −3 or x = 8

So y is negative on the interval (−3,8) and positive on the intervals (−∞,−3) and (8,∞)

y = x² + 4x + 12 can't be factored and you will find, based on the quadratic formula, that it has no zeroes (at least not any that are in the set of real numbers).

You know there are no zeroes when you calculate b² − 4ac = (4)² − 4(1)(12) = 16 − 48 < 0.

b² − 4ac ≥ 0 is required for the quadratic function to have zeroes

(one zero if b² − 4ac = 0; two zeroes if b² − 4ac > 0).

since the function y = x² + 4x + 12 has no zeroes that means it is never crosses the x-axis. This means that it is positive for all values of x.

So y is positive on the interval (−∞,∞)

y = 5x² − 3x – 8 = (x + 1)(5x – 8) so y = 0 when x = −1 or x = 8/5

So y is negative on the interval (−1,8/5) and positive on the intervals (−∞,−1) and (8/5,∞)

5x² − 3x – 8 is a little bit more challenging to factor (because a = 5 instead of 1).

But you can always find the same zeroes using the quadratic formula if you get stuck in the factoring step.

Answer 2

y=x²+2x-8 Factor this to find the zeros

y=(x+4)(x-2) Set the factors equal to zero

x+4=0 and x-2=0

x=-4 and x=2 are the zeros. These 2 zeros split the number line into 3 parts. Pick a test point in each of the 3 sections and evaluate at these values. I just need to sign to see if it is positive or negative

x=-5 (-5+4)(-5-2) = (negative)(negative) = positive

x=0 (0+4)(0-2) = (positive)(negative) = negative

x=3 (3+4)(3-2) = (positive)(positive) = positive

(-∞,-4) U (2,∞)