Question

The function h(x) is quadratic and h(3) = h(–10) = 0. Which could represent h(x)?

Answer 1

I hope this helps you

Answer 2

Answer:

$h(x)=2x^2+14x-60$

Step-by-step explanation:

This question can be solved by two methods

Method 1: Substitute x=3 and x=-10 in all the equations and determine which equals to zero (ie., check h(3)=0 and h(-10)=0 for all the equations)

Equation 1

$h(x)=x^2-13x-30$

$h(3)=3^2-13(3)-30$

$h(3)=-60$

As h(3)≠0, Equation 1 is discounted

Equation 2

$h(x)=x^2-7x-30$

$h(3)=3^2-7(3)-30$

$h(3)=-42$

As h(3)≠0, Equation 2 is discounted

Equation 3

$h(x)=2x^2+26x-60$

$h(3)=2(3)^2+26(3)-60$

$h(3)=36$

As h(3)≠0, Equation 3 is discounted

Equation 4

$h(x)=2x^2+14x-60$

$h(3)=2(3)^2+14(3)-60$

$h(3)=0$

$h(x)=2x^2+14x-60$

$h(-10)=2(-10)^2+14(-10)-60$

$h(-10)=0$

As h(3)=0 and h(-10)=0, Equation 4 represents h(x)

Method 2: Solve to find the roots of each equation where h(x)=0 using the quadratic formula. Roots should be x=3,x=-10

The quadratic formula is:

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

where a, b and c are as below

$h(x)=ax^2+bx+c=0$

Equation 1

$h(x)=x^2-13x-30=0$

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

$x=\frac{13\±\sqrt{(-13)^2-4(1)(-30)}}{2(1)}$

$x=15,x=-2$

As roots are not x=3 and x=-10, Equation 1 is discounted

Equation 2

$h(x)=x^2-7x-30$

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-7)\±\sqrt{(-7)^2-4(1)(-30)}}{2(1)}$

$x=10,x=-3$

As roots are not x=3 and x=-10, Equation 2 is discounted

Equation 3

$h(x)=2x^2+26x-60$

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(26)\±\sqrt{(26)^2-4(2)(-60)}}{2(2)}$

$x=2,x=-15$

As roots are not x=3 and x=-10, Equation 3 is discounted

Equation 4

$h(x)=2x^2+14x-60$

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(14)\±\sqrt{(14)^2-4(2)(-60)}}{2(2)}$

$x=3,x=-10$

As roots are x=3 and x=-10, Equation 4 represents h(x)