Question

A cubic polynomial function f has a leading coefficient 2 and a constant term 7 if f(1)=7 and f(2)=9 what is f(-2)?

Answer 1

Steps:

So what we know with this function is that:

• Points on this function are (1,7) and (2,9)
• The leading term is 2x³ (2 is the leading coefficient, and cubic = degree of 3)
• The constant is 7.
• Additionally, remember that a cubic function template is $y=ax^3+bx^2+cx+d$

Using this info, we can create 2 different equations:

$7=2*1^3+b*1^2+c*1+7\\9=2*2^3+b*2^2+c*2+7$

Now, simplify each of them:

$7=2*1^3+b*1^2+c*1+7\\7=2+b+c+7\\7=9+b+c\\-2=b+c\\\\9=2*2^3+b*2^2+c*2+7\\9=2*8+4b+2c+7\\9=16+4b+2c+7\\9=23+4b+2c\\-14=4b+2c$

Now with these 2 simplified equations, we can do a system of equations to solve for b and c of this cubic function. For this, I will be using the elimination method. Firstly, divide both sides of the second equation by 2:

$-2=b+c\\-7=2b+c$

Next, subtract the first equation from the second equation to get $-5=b$ . We have found b.

Next, to find c plug in -5 into the b variable either equation and then solve for c:

$-2=-5+c\\3=c\\\\-14=4(-5)+2c\\-14=-20+2c\\6=2c\\3=c$

Putting it together, our cubic function is $f(x)=2x^3-5x^2+3x+7$

Now, to find f(-2), plug -2 into the x variable and solve:

$f(-2)=2(-2)^3-5(-2)^2+3(-2)+7\\f(-2)=2*(-8)-5*4-6+7\\f(-2)=-16-20-6+7\\f(-2)=-16-20-6+7\\f(-2)=-35$

Answer:

In short, f(-2) = -35.

Answer 2

A Cubic polynomial function can be defined as a polynomial function with highest degree of its exponents as 3. The standard form of a cubic polynomial function is:

$\rm ax^3 + bx^2 + cx + d = 0$, where $\rm a \neq 0$ and d is constant.

For the given question the cubic equation will be:

$\rm 2x^3 + bx^2 + cx + 7 = 0$

The values of b and c is -5 and 3 respectively.

The value of $\rm f(-2) = -35$

To reach the above answers, following calculation is required:

Given:

$\begin{aligned} \rm Leading\: coefficient\:(a) &= 2\\Constant \:Term(d) &= 7\\f(1) &= 7\\f(2) &= 9 \end$

Therefore the equation formed will be:

$\rm 2x^3 + bx^2 + cx + 7 = 0$

On solving $\rm f(1) = 7$

$\begin{aligned} \rm 2(1)^3 + b(1)^2 + c(1) + 7 &= 7\\2 + b + c + 7 &= 7\\9 + b + c &= 7\\b + c &= 7 - 9\\b + c &= -2 \\b &= -2 -c\end$                           ...(1)

Solving $\rm f(2) = 9$

$\begin{aligned} \rm 2(2)^3 + b(2)^2 + c(2) + 7 &= 9\\2(8) + b(4) + c(2) + 7 &= 9\\16 + 4b + 2c + 7 &= 9\\23 + 4b + 2c &= 9\\4b + 2c &= 9 - 23\\4b + 2 c &= -14 \end$                         ...(2)

Substituting value of b from equation 1 in equation 2:

$\begin{aligned} \rm 4(-2 -c) + 2c &= -14\\-8 -4c + 2c &= -14\\-8 -2c &= -14\\-2c &= -14 + 8\\-2c &= -6\\c &= \dfrac{-6}{-2} \\c &= 3 \end$

Substituting c=3 in equation 1:

$\rm \begin{aligned} b &= -2 -c\\b &= -2 -3\\b &= -5 \end$

On substituting values of b and c, the cubic equation will be:

$\rm 2x^3 -5x^2 + 3x + 7 = 0$

Calculating $\rm f(-2)$

$= 2(-2)^3 -5(-2)^2 +3(-2) + 7\\= 2(-8) -5(4) -6 + 7\\= -16 -20 -6 + 7\\= -35$

Therefore the solution to $\rm f(-2) = -35$.

Learn more about cubic equations here:

https://brainly.com/question/1727936