To determine the zeros of the quadratic function \( y = 2x^2 + 9x + 4 \), we need to find the values of \( x \) that make \( y = 0 \).
This involves solving the quadratic equation:
\[ 2x^2 + 9x + 4 = 0 \]
We can use the quadratic formula to find the roots, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation:
- \( a = 2 \)
- \( b = 9 \)
- \( c = 4 \)
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 9^2 - 4 \cdot 2 \cdot 4 = 81 - 32 = 49 \]
Since the discriminant is positive, there are two distinct real roots.
Next, we calculate the roots using the quadratic formula:
\[ x = \frac{-9 \pm \sqrt{49}}{2 \cdot 2} \]
Calculating \( \sqrt{49} \):
\[ \sqrt{49} = 7 \]
Now, substituting this value back into the formula, we get:
\[ x = \frac{-9 \pm 7}{4} \]
Calculating the two potential solutions:
- For \( -9 + 7 = -2 \):
\[ x_1 = \frac{-2}{4} = -0.5 \]
- For \( -9 - 7 = -16 \):
\[ x_2 = \frac{-16}{4} = -4 \]
Thus, the zeros of the function \( y = 2x^2 + 9x + 4 \) are:
\[ (-4, 0) \quad \text{and} \quad (-0.5, 0) \]
So the correct response is:
(−4,0)(−0.5,0)