Question

Find b and c so that the parábola y=-3x^2 +bx+c has x-intercepts 1 and -4

Answer 1

Answer:

so b=-9 while c=12.

Step-by-step explanation:

If you have x-intercepts 1 and -4, then that means f(1)=0 and f(-4)=0.

You are given $f(x)=-3x^2+bx+c$

So you have the system of equations to solve:

$f(1)=-3(1)^2+b(1)+c=0$

$f(-4)=-3(-4)^2+b(-4)+c=0$

Evaluating the exponents:

$-3(1)+b+c=0$

$-3(16)-4b+c=0$

Doing a little bit of multiplying:

$-3+b+c=0$

$-48-4b+c=0$

Let's add 3 on both sides of equation 1 and 48 on both sides of equation 2:

$b+c=3$

$-4b+c=48$

Subtracting the equations will eliminate c.

Let's do that:

$5b+0c=-45$

$5b=-45$

Divide both sides by 5:

$b=\frac{-45}{5}$

Simplify:

$b=-9$

If b=-9 and b+c=3 then -9+c=3 implies c=9+3=12.

so b=-9 while c=12.