Question #8138e

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Answer 1 · 2016-02-28T11:33:02

$a = 7; b = 1$ $a=7" ; "b=1$

Given: $h (x) \to x^{2} + 1 k (x) \to a x + b$ $h(x)-> x^2+1" "k(x)-> ax+b$

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

$h (0) = k (0) \to {(0)}^{2} + 1 = a (0) + b$ $h(0) =k(0) -> (0)^2+1 = a(0)+b$

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

Now we have: $h (x) \to x^{2} + 1 k (x) \to a x + 1$ $h(x)-> x^2+1" "k(x)-> ax+1$

$k (2) \to 15 = a x + 1$ $k(2)-> 15=ax+1$

$a x = 14$ $ax=14$

$a (2) = 14$ $a(2)=14$

$a = 7$ $a=7$