- Set the quadratic function \(y = x^2 + 7x - 30\) to zero.
- Factor the expression as \((x - 3)(x + 10) = 0\).
- Solve for \(x\) by setting each factor to zero, yielding \(x = 3\) and \(x = -10\).
- The final zeros are \(\boxed{3, -10}\).
Explanation
- Analyzing the Problem
To find the zeros of the quadratic function \(y = x^2 + 7x - 30\), we set the function equal to zero:
\[x^2 + 7x - 30 = 0\]
A zero of a function is a value of \(x\) for which the output \(y\) is zero. We can solve this quadratic equation by factoring or by using the quadratic formula.
- Factoring the Quadratic Equation
We look for two numbers that multiply to the constant term \(-30\) and add to the coefficient of the linear term \(7\).
Listing the pairs of factors of \(-30\):
- \(-1 \times 30 = -30\), sum = \(29\)
- \(-2 \times 15 = -30\), sum = \(13\)
- \(-3 \times 10 = -30\), sum = \(7\)
Since the numbers \(-3\) and \(10\) satisfy both conditions, we can factor the quadratic expression as:
\[(x - 3)(x + 10) = 0\]
- Solving and Verifying the Zeros
Setting each factor equal to zero, we solve for \(x\):
- \(x - 3 = 0 \implies x = 3\)
- \(x + 10 = 0 \implies x = -10\)
Thus, the zeros of the function are \(x = 3\) and \(x = -10\). We verify these values by substituting them back into the original equation:
- For \(x = 3\): \((3)^2 + 7(3) - 30 = 9 + 21 - 30 = 0\).
- For \(x = -10\): \((-10)^2 + 7(-10) - 30 = 100 - 70 - 30 = 0\).
Both solutions are verified. The zeros of the function are \(\boxed{3, -10}\).
Examples
Quadratic functions like this are used in physics to model projectile motion. For example, if a ball is thrown into the air, its height over time might be described by a quadratic function. Finding the zeros of this function helps determine when the object hits the ground, which is essential for predicting the trajectory and landing point of the projectile in various sports and engineering applications.