Question

which function has real zeros at x = −10 and x = −6? f(x) = x2 16x 60 f(x) = x2 − 16x 60 f(x) = x2 4x 60 f(x) = x2 − 4x 60

Answer 1

Answer:

Option A is correct

The function $x^2+16x+60$ has real zeroes at x =-10 and x =-6

Explanation:

Given: The real zeroes or roots are x = -10, and x = -6

To find the quadratic function of degree 2.

$x^2- (\alpha+\beta)x + \alpha\beta =0$ where α,β are real roots.   ....[1]

Here, α= -10  and β= -6

Sum of the roots:

α+β =  -10+(-6) = -10-6 = -16

Product of the roots:

αβ = (-10)(-6)= 60

Substitute these value in equation [1] we have;

$x^2-(-16)x+60 = x^2+16x+60$

Therefore, the quadratic function for the real roots at x =-10 and x =-6 ;

$x^2+16x+60$

Answer 2

Would assume the function is a quadratic function. It is degree 2.

The zeros or roots are x = - 10, and x = -6.

For quadratic:

x² - (sum of roots)x + product = 0

x² - (-10 + -6)x + (-10)*(-6) = 0

x² - (-16)x + (60) = 0

x² + 16x + 60 = 0

f(x) = x² + 16x + 60