To find the zeros of the quadratic function \( y = 2x^2 + 9x + 4 \), we need to make the equation equal to zero and solve for \( x \):
\[ 2x^2 + 9x + 4 = 0 \]
We can use the quadratic formula to find \( x \). The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, the numbers are:
- \( a = 2 \)
- \( b = 9 \)
- \( c = 4 \)
First, let's find the part called the discriminant, which is \( b^2 - 4ac \):
\[ b^2 - 4ac = 9^2 - 4 \times 2 \times 4 \] \[ = 81 - 32 \] \[ = 49 \]
Now we can put our numbers into the quadratic formula:
\[ x = \frac{-9 \pm \sqrt{49}}{2 \cdot 2} \] \[ = \frac{-9 \pm 7}{4} \]
Next, we'll find the two possible values for \( x \):
-
For \( -9 + 7 \): \[ x = \frac{-2}{4} = -\frac{1}{2} \]
-
For \( -9 - 7 \): \[ x = \frac{-16}{4} = -4 \]
So the zeros of the quadratic function \( y = 2x^2 + 9x + 4 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -4 \]
That means the points where the graph crosses the x-axis are at \( x = -\frac{1}{2} \) and \( x = -4 \).