Question

What are the zeros of the graph of: y = 2x^2 + 5x - 12

Answer 1

To find the zeros of the function \( y = 2x^2 + 5x - 12 \), we need to solve the equation \( 2x^2 + 5x - 12 = 0 \) for \( x \).

We can use the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the quadratic \( ax^2 + bx + c \), we have:

- \( a = 2 \)
- \( b = 5 \)
- \( c = -12 \)

Now, we can calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 5^2 - 4(2)(-12) = 25 + 96 = 121
\]

Since the discriminant is positive, we will have two distinct real solutions.

Now we can substitute these values into the quadratic formula:

\[
x = \frac{-5 \pm \sqrt{121}}{2 \cdot 2}
\]

Calculating \( \sqrt{121} = 11 \):

\[
x = \frac{-5 \pm 11}{4}
\]

This gives us two potential solutions:

1. For the plus sign:

\[
x = \frac{-5 + 11}{4} = \frac{6}{4} = \frac{3}{2}
\]

2. For the minus sign:

\[
x = \frac{-5 - 11}{4} = \frac{-16}{4} = -4
\]

Thus, the zeros of the graph \( y = 2x^2 + 5x - 12 \) are:

\[
x = \frac{3}{2} \quad \text{and} \quad x = -4
\]