Question 1097183
{{{h(x)=ax+b}}}
So then,
{{{h(-3)=a(-3)+b=3}}}
1.{{{-3a+b=3}}}
and
{{{h(0)=a(0)+b=2}}}
2.{{{b=2}}}
Substituting into eq. 1,
{{{-3a+2=3}}}
{{{-3a=1}}}
{{{a=-1/3}}}
So,
{{{h(x)=-x/3+2}}}
When {{{x=-6}}},
{{{h(-6)=-(-6)/3+2=2+2=4}}}
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*[illustration CK5.JPG].