Q. Which Function Has Zeros at \( x = -2 \) and \( x = 5 \)?

Q: How can I write the function in standard form?

Expand (x + 2)(x - 5) to get x^2 - 3x - 10. For general scale, write f(x) = kx^2 - 3kx - 10k.

Q: If I want integer coefficients, what choices of k work?

Any integer k gives integer coefficients. The simplest is k = 1, yielding x^2 - 3x - 10.

Q: How does the graph behave near x = -2 and x = 5?

If each root has multiplicity 1, the graph crosses the x-axis linearly at both points. If multiplicity is even, the graph touches and turns around at that root.

Q: Can non-polynomial functions have these zeros?

Yes. Rational functions, trigonometric polynomials, or piecewise functions can have x = -2 and x = 5 as zeros, but zeros are typically expressed vifactors in polynomial factors for polynomials.

Answer

Write the linear factors.
Zeros at -2 and 5 mean the factors are (x + 2) and (x – 5).
Build the polynomial.
The general form is f(x) = k(x + 2)(x – 5).
Expand the simplest choice.
Let k = 1 and multiply.

\[ f(x) = x^2 – 3x – 10 \]

Detailed Explanation

Solution — Find a function with zeros at x = -2 and x = 5

Understand what a zero means.

A real number r is a zero (root) of a function f if f(r) = 0. For a polynomial, each zero r corresponds to a linear factor of the form \(x – r\).
Write the linear factors for the given zeros.

For the zero \(x = -2\) the corresponding factor is \(x – (-2) = x + 2\). For the zero \(x = 5\) the corresponding factor is \(x – 5\).
Form the polynomial by multiplying the factors.

The polynomial having those zeros (up to a nonzero constant factor) is
\[
f(x) = a\,(x + 2)(x – 5),
\]
where \(a\) is any nonzero constant. Choosing \(a = 1\) gives the simplest monic polynomial.
Expand the product (optional, to write as a standard quadratic).

Multiply the factors:
\[
(x + 2)(x – 5) = x^2 – 5x + 2x – 10 = x^2 – 3x – 10.
\]
So with \(a = 1\),
\[
f(x) = x^2 – 3x – 10.
\]
Verify the zeros (optional check).

Evaluate at \(x = -2\):
\[
f(-2) = (-2)^2 – 3(-2) – 10 = 4 + 6 – 10 = 0.
\]
Evaluate at \(x = 5\):
\[
f(5) = 5^2 – 3(5) – 10 = 25 – 15 – 10 = 0.
\]
Both give zero, confirming the function has the required zeros.

Final answer: The family of functions with zeros at \(x = -2\) and \(x = 5\) is

\[
f(x) = a\,(x + 2)(x – 5),
\]

where \(a\) is any nonzero constant. The simplest choice is \(f(x) = x^2 – 3x – 10\).

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Frequently Asked Questions

Which simplest polynomial has zeros at x = -2 and x = 5?

f(x) = (x + 2)(x - 5). Expanded: f(x) = x^2 - 3x - 10. This is the monic quadratic with those zeros.

Are there other functions with those same zeros?

Yes — infinitely many. Multiply (x + 2)(x - 5) by any nonzero constant or by additional factors that include those factors with multiplicity to get other polynomials.

How do I get polynomial with those zeros and specific y-intercept?

Multiply (x + 2)(x - 5) by constant k, then choose k so f(0) equals the desired y-intercept: k = (desired y-intercept)/((0+2)(0-5)).

What does multiplicity mean for these zeros?

Multiplicity is how many times factor repeats. If (x + 2) or (x - 5) appears squared, the graph touches the axis (even multiplicity) instead of crossing (odd multiplicity).

How does the leading coefficient affect the graph?

The leading coefficient (k) scales the graph vertically and can flip it if negative; it does not change the zeros. For quadratic k>0 opens up, k<0 opens down.

How can I write the function in standard form?

Expand (x + 2)(x - 5) to get x^2 - 3x - 10. For general scale, write f(x) = kx^2 - 3kx - 10k.

If I want integer coefficients, what choices of k work?