To find the zeros of the quadratic function \( y = 4x^2 + 9x - 9 \), we need to solve the equation:
\[ 4x^2 + 9x - 9 = 0 \]
We can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, the coefficients are:
- \( a = 4 \)
- \( b = 9 \)
- \( c = -9 \)
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 9^2 - 4(4)(-9) = 81 + 144 = 225 \]
Since the discriminant is positive, there will be two real and distinct solutions. Next, we apply the quadratic formula:
\[ x = \frac{-9 \pm \sqrt{225}}{2 \cdot 4} \]
Calculating \( \sqrt{225} \):
\[ \sqrt{225} = 15 \]
Now substituting back into the formula:
\[ x = \frac{-9 \pm 15}{8} \]
This gives us two possible solutions:
-
\( x = \frac{-9 + 15}{8} = \frac{6}{8} = \frac{3}{4} \)
-
\( x = \frac{-9 - 15}{8} = \frac{-24}{8} = -3 \)
Thus, the zeros of the graph of the function \( y = 4x^2 + 9x - 9 \) are:
\[ x = \frac{3}{4} \quad \text{and} \quad x = -3 \]