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:
- For the plus sign:
\[ x = \frac{-5 + 11}{4} = \frac{6}{4} = \frac{3}{2} \]
- 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:
\[ x = \frac{3}{2} \quad \text{and} \quad x = -4 \]