Domain Analysis of y=-4x²+x+3: Finding Positive and Negative Regions

Find the positive and negative domains of the following function:

$y=-4x^2+x+3$

To solve this problem, let's start by finding the roots of the quadratic equation:

The given function is $y = -4x^2 + x + 3$. We set it to zero to find the roots:

$-4x^2 + x + 3 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = -4$, $b = 1$, and $c = 3$:

$x = \frac{-1 \pm \sqrt{1^2 - 4(-4)(3)}}{2(-4)}$

$x = \frac{-1 \pm \sqrt{1 + 48}}{-8}$

$x = \frac{-1 \pm \sqrt{49}}{-8}$

$x = \frac{-1 \pm 7}{-8}$

The roots are:

• $x_1 = \frac{-1 + 7}{-8} = \frac{6}{-8} = -\frac{3}{4}$
• $x_2 = \frac{-1 - 7}{-8} = \frac{-8}{-8} = 1$

These roots divide the number line into intervals: $(-\infty, -\frac{3}{4})$, $(-\frac{3}{4}, 1)$, and $(1, \infty)$.

We test each interval to determine where the function is positive or negative:

Interval $(-\infty, -\frac{3}{4})$: Choose $x = -1$.

• $y = -4(-1)^2 + (-1) + 3 = -4 - 1 + 3 = -2$ (Negative)

Interval $(-\frac{3}{4}, 1)$: Choose $x = 0$.

• $y = -4(0)^2 + 0 + 3 = 3$ (Positive)

Interval $(1, \infty)$: Choose $x = 2$.

• $y = -4(2)^2 + 2 + 3 = -16 + 2 + 3 = -11$ (Negative)

Therefore, the function is positive in the interval $(-\frac{3}{4}, 1)$ and negative in the intervals $(-\infty, -\frac{3}{4})$ and $(1, \infty)$.

Thus, the positive and negative domains of the function are:

$x > 0 : -\frac{3}{4} < x < 1$

$x > 1$ or $x < 0 : x < -\frac{3}{4}$

The correct answer choice corresponds to:

$x > 1$ or $x<0:x<-\frac{3}{4}$

$x > 0 : -\frac{3}{4} < x <1$