Question

write a quadratic function h whose zeros are -3 and 13.
h(x) =

Explanation:

Step1: Recall the factor theorem

If a quadratic function has zeros \( r_1 \) and \( r_2 \), then it can be written in factored form as \( h(x)=a(x - r_1)(x - r_2) \), where \( a \) is a non - zero constant. For simplicity, we can take \( a = 1 \) (we can choose any non - zero value for \( a \), and choosing \( a=1 \) gives a simple quadratic function). Here, \( r_1=-3 \) and \( r_2 = 13 \).

Step2: Substitute the zeros into the factored form

Substitute \( r_1=-3 \) and \( r_2 = 13 \) into the formula \( h(x)=a(x - r_1)(x - r_2) \). When \( a = 1 \), we have \( h(x)=(x-(-3))(x - 13)=(x + 3)(x - 13) \).

Step3: Expand the factored form

We use the FOIL method to expand \( (x + 3)(x - 13) \).
First terms: \( x\times x=x^{2} \)
Outer terms: \( x\times(-13)=-13x \)
Inner terms: \( 3\times x = 3x \)
Last terms: \( 3\times(-13)=-39 \)

Then combine like terms: \( x^{2}-13x + 3x-39=x^{2}-10x - 39 \)

Answer:

\( h(x)=x^{2}-10x - 39 \) (or any non - zero multiple of this, for example, if \( a = 2 \), \( h(x)=2(x + 3)(x - 13)=2(x^{2}-10x - 39)=2x^{2}-20x - 78 \), but the simplest form with \( a = 1 \) is \( x^{2}-10x - 39 \))