Quadratic Function Domain Analysis: Positive and Negative Regions in y = 4x² - x - 3

Find the positive and negative domains of the following function:

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

To solve for the positive and negative domains of the function $y = 4x^2 - x - 3$, follow these steps:

• Step 1: Identify and apply the quadratic formula to find the roots.
• Step 2: Calculate the discriminant to ensure real roots.
• Step 3: Analyze the sign changes in the resulting intervals.

Step 1: The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = 4$, $b = -1$, and $c = -3$.

Step 2: Calculate the discriminant: $b^2 - 4ac = (-1)^2 - 4 \times 4 \times (-3) = 1 + 48 = 49$. Since the discriminant is positive, two distinct real roots exist.

Step 3: Calculate the roots using the formula:

$x = \frac{-(-1) \pm \sqrt{49}}{2 \times 4} = \frac{1 \pm 7}{8}$

Thus, the roots $x_1 = \frac{1 + 7}{8} = 1$ and $x_2 = \frac{1 - 7}{8} = -\frac{3}{4}$.

Now we examine the sign of $y = 4x^2 - x - 3$ across the intervals determined by these roots: $(-\infty, -\frac{3}{4})$, $(-\frac{3}{4}, 1)$, and $(1, \infty)$.

• For $x < -\frac{3}{4}$: Choose $x = -1$. Substitute into the function: $y = 4(-1)^2 - (-1) - 3 = 4 + 1 - 3 = 2$. Positive.
• For $-\frac{3}{4} < x < 1$: Choose $x = 0$. Substitute: $y = 4(0)^2 - (0) - 3 = -3$. Negative.
• For $x > 1$: Choose $x = 2$. Substitute: $y = 4(2)^2 - (2) - 3 = 16 - 2 - 3 = 11$. Positive.

Therefore, the positive domains are $x < -\frac{3}{4}$ and $x > 1$, and the negative domain is $-\frac{3}{4} < x < 1$.

The positive and negative domains are: $x < 0 : -\frac{3}{4} < x < 1$ and $x > 1$ or $x > 0 : x < -\frac{3}{4}$.