To locate the zeros of the quadratic equation \( y = -3x^2 + 12x + 15 \), we can use the quadratic formula, factoring, or completing the square. Let's first use the quadratic formula:
The quadratic formula states that for any quadratic equation of the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) can be found using:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, the coefficients are:
- \( a = -3 \)
- \( b = 12 \)
- \( c = 15 \)
Step 1: Calculate the Discriminant
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 12^2 = 144 \] \[ 4ac = 4 \times (-3) \times 15 = -180 \] \[ b^2 - 4ac = 144 - (-180) = 144 + 180 = 324 \]
Step 2: Apply the Quadratic Formula
Now, substituting the values into the quadratic formula:
\[ x = \frac{-12 \pm \sqrt{324}}{2 \times -3} \]
Calculating \( \sqrt{324} \):
\[ \sqrt{324} = 18 \]
Substituting back into our formula:
\[ x = \frac{-12 \pm 18}{-6} \]
This gives us two possible solutions:
-
Using the positive root: \[ x = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1 \]
-
Using the negative root: \[ x = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5 \]
Step 3: Solutions
Therefore, the zeros of the equation \( y = -3x^2 + 12x + 15 \) are:
\[ x = -1 \quad \text{and} \quad x = 5 \]
These are the points where the quadratic function intersects the x-axis.