Question

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

h(x) = x2 – 13x – 30
h(x) = x2 – 7x – 30
h(x) = 2x2 + 26x – 60
h(x) = 2x2 + 14x – 60

Answer 1

I hope this helps you

Answer 2

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

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=(-b\±√(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=(-b\±√(b^2-4ac))/(2a)`

`x=(13\±√((-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=(-b\±√(b^2-4ac))/(2a)`

`x=(-(-7)\±√((-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=(-b\±√(b^2-4ac))/(2a)`

`x=(-(26)\±√((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=(-b\±√(b^2-4ac))/(2a)`

`x=(-(14)\±√((14)^2-4(2)(-60)))/(2(2))`

`x=3,x=-10`

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