The function is$$h(x)=egincasesdfrac4x^2+5x-6x+2 & extif $x eq-2$, \<6pt>3x+k & extif $x= -2$. \endcases$$

After factoring the fraction I to be left through $4x-3= -11$ as soon as $x=-2$. If i multiply $(x+2)(4x-3)$ and then plug $-2$ right into the resulting equation ns am left through $-24$. If I do $4(-2)-3=3(-2)+k$ i am left v $k=-5$ which is none of the answer choices.

You are watching: Find the value of k that makes the function continuous

The answer choices to this inquiry are

A. $k=20$

B. $k=0$

C. $k=2$

D. $k=4$

I am not sure where i go from below so any helpful tips on where I must start following would it is in appreciated.

calculus algebra-precalculus boundaries
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edited Jun 5 "16 in ~ 21:14

will R
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inquiry Jun 5 "16 at 20:57

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2 answers 2

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The right value is $k=-5$ as you have actually found.

Observing that, together $x eq-2$, friend have$$f(x)=frac4x^2+5x-6x+2=frac(4x-3)(x+2)x+2=4x-3$$ i beg your pardon is same to $-11$ together $x o -2$. At the same time putting $x=-2$ in $3x+k$ provides $-6+k$ climate you will require that$$-6+k=-11$$ that is $k=-5$.

edited Jun 5 "16 at 21:06
answered Jun 5 "16 in ~ 21:01

Olivier OloaOlivier Oloa
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Using the dominion of l"Hospital top top $lim_x o -2frac4x^2+5x-6x+2$ we acquire $lim_x o -28x+5=-11$. Because that none the the given alternatives it is feasible that $f$ is continuous since $-6+k eq -11$ because that $kin,2,4,20$.

reply Jun 5 "16 in ~ 21:08

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