Question #8138e

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Answer 1 · 2016-02-28T11:33:02

$a=7" ; "b=1$

Given: $h(x)-> x^2+1" "k(x)-> ax+b$

'~~~~~~~~~~~~~~~~~~~~~~~
Condition 1

$h(0) =k(0) -> (0)^2+1 = a(0)+b$

$=> b=+1$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Condition 2

Now we have: $h(x)-> x^2+1" "k(x)-> ax+1$

$k(2)-> 15=ax+1$

$ax=14$

$a(2)=14$

$a=7$