Finding constants in a quadratic equation
By Sarah Smith
I have the problem: Given that $g(x)$ is a quadratic function; that is in the form $g(x) = ax^2+bx+c$ for some constants $a,b,c$. Find the values of the constants $a,b,c$ given that $g(1)=6$, $g'(2)=24$, and $g''(3)=-4$.
I'm confused because it doesn't give me a the value for $g(0)$ so I don't know how i would start this problem.
$\endgroup$ 32 Answers
$\begingroup$You have three equations with three unknowns. Take the first derivative to get
$$g'(x)=2ax+b$$
and the second derivative
$$g''(x)=2a$$.
Now, we know that the second derivative at $x=3$ is -4. So,
$$g''(3)=2a=-4$$,
therefore, $a=-2$.
Now, we know that $g'(2)=24$, so
$$-4(2)+b=24$$ $$b=32.$$
Finally, we know that $g(1)=6$, so
$$-2(1)^2+32+c=6,$$
$$30+c=6$$
$$c=-24$$.
$\endgroup$ $\begingroup$You have $g(1) = 6$. That means that $$ a 1^2 + b 1 + c = 6 $$ i.e., $$ a + b + c = 6. (*) $$ You have $g'(2) = 24$, which means that that $$ 2a\cdot 2 + b = 24 $$ i.e. $$ 4a + b= 24 (**) $$ From the second derivative fact, you can derive a third equation: $$ 2a = -4 (***) $$
Now, using the three starred equations, working from the bottom up, you can find $a$, $b$, and $c$.
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