The polynomial f(x) has degree 3. If f(-1) = 15, f(0) = 0, f(1) = -5, and f(2) = 12, then what are the x-intercepts of the graph of f?

Thanks so much,

PartialMathematician
Dec 7, 2018

#1**+5 **

I have this so far:

From the problem, we know that $x^3$ is the leading term, and $3$ is the largest power. We also know that $f(x)$ has no constant because $f(0) = 0$. We can write our equation as $f(x) = a_3x^3 + a_2x^2 + a_1x$ with $a_3, a_2,$ and $a_1$ as coefficients of different powers of $x$. We can also factor $f(x) = a_3x^3 + a_2x^2 + a_1x$ into $f(x) = x(a_3x^2 + a_2x + a_1)$.

With the information $f(1) = -5$, we know that $a_3 + a_2 + a_1 = -5$ since $1^n = 1$.

Using the information $f(2) = 12$, we can plug in $2$ for $x$ in $f(x) = x(a_3x^2 + a_2x + a_1)$ to get $12 = 2(a_3(2^2) + a_2(2) + a_1)$. The equation simplifies into $12 = 2(4a_3 + 2a_2 + a_1)$, and by dividing both sides by $2$, we have $4a_3 + 2a_2 + a_1 = 12$.

Using the information $f(-1) = 15$, we can plug in $-1$ for $x$ in $f(x) = x(a_3x^2 + a_2x + a_1)$ to get $15 = -(a_3((-1)^2) + a_2(-1) + a_1)$. The equation simplifies into $15 = -(a_3 - a_2 + a_1)$, and by multiplying both sides by $-1$, we have $a_3 - a_2 + a_1 = -15$.

We have the three equations [#1] $a_3 + a_2 + a_1 = -5$, [#2] $4a_3 + 2a_2 + a_1 = 12$, and [#3] $a_3 - a_2 + a_1 = -15$. Using system of equations on the first and thrid equations, we have $(a_3 + a_2 + a_1) - (a_3 - a_2 + a_1) = -5 - (-15)$, which simplifies into $2a_2 = 10 \Rightarrow a_2 = 5$. Substituting $a_2 = 5$ in the first equation, we have [#4] $a_3 + a_1 = -10$, and substituting $a_2 = 5$ in the second equation, we have [#5] $4a_3 + a_1 = 2$. Using system of equations on the fourth and fifth equations, we have $(a_3 + a_1) - (4a_3 + a_1) = -10-2$, which simplifies into $-3a_3 = -12 \Rightarrow a_3 = 4$. Plugging in the values $a_2 = 5$ and $a_3 = 4$ into the first equation, we have $4 + 5 + a_1 = -5$, which simplifies to $a_1 = -14$.

Plugging the values $a_2 = 5$, $a_3 = 4$, and $a_1 = -14$ into $f(x) = x(a_3x^2 + a_2x + a_1)$, we have $f(x) = x(4x^2 + 5x -14)$.

PartialMathematician
Dec 7, 2018

#2**+4 **

Here is the continuation:

Plugging the values $a_2 = 5$, $a_3 = 4$, and $a_1 = -14$ into $f(x) = x(a_3x^2 + a_2x + a_1)$, we have $f(x) = x(4x^2 + 5x -14)$. Since $y = f(x)$ and the x-intercept is found when $y = 0$, we can rewrite our equation as $x(4x^2 + 5x -14) = 0$. Using the quadratic formula $x = \dfrac{-b\pm \sqrt{b^2 - 4ac}}{2a}$ with $a=4$, $b=5$, and $c=-14$, we have $x = \dfrac{-5\pm \sqrt{5^2 - 4(4)(-14)}}{2(4)}$, which simplifies into $x = \dfrac{-5\pm \sqrt{249}}{8}$.

Revisiting the equation $x(4x^2 + 5x -14) = 0$, we now know that $x(4x^2 + 5x -14) = 0$ when $x = 0$, $x = \dfrac{-5 + \sqrt{249}}{8}$, or $x = \dfrac{-5 - \sqrt{249}}{8}$. These are the three x-intercepts of the graph of $f$: $\boxed{0}$, $\boxed{\dfrac{-5 + \sqrt{249}}{8}}$, and $\boxed{\dfrac{-5 - \sqrt{249}}{8}}$. If these x-intercepts are written in decimals, they would be $0$, $1.347$, and $-2.597$.

PartialMathematician
Dec 7, 2018

#7**+6 **

**The polynomial f(x) has degree 3. If f(-1) = 15, f(0) = 0, f(1) = -5, and f(2) = 12,**

** then what are the x-intercepts of the graph of f?**

\(\boxed{f(x) =ax^3+bx^2+cx+d}\) degree 3

\(\begin{array}{|lrcll|} \hline f(0) = 0: & f(0) = 0 &=& a\cdot 0^3+ b\cdot 0^2+ c\cdot 0 +d \\ & 0 &=& d \\ & \mathbf {d }& \mathbf{=}& \mathbf{0} \\ \hline \end{array} \)

So \(\boxed{f(x) =ax^3+bx^2+cx}\)

\(\begin{array}{|lrcll|} \hline f(1) = -5: & f(1) = -5 &=& a\cdot 1^3+ b\cdot 1^2+ c\cdot 1 \\ & -5 &=& a + b + c \\ & \mathbf {a + b + c }& \mathbf{=}& \mathbf{-5} \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline f(-1) = 15: & f(-1) = 15 &=& a\cdot (-1)^3+ b\cdot (-1)^2+ c\cdot (-1) \\ & 15 &=& -a + b - c \\ & \mathbf {-a + b - c }& \mathbf{=}& \mathbf{15}\qquad (2) \\ \hline \end{array}\)

**\(\mathbf{(1)+(2):}\)**

\(\begin{array}{|lrcll|} \hline (1) & \mathbf {a + b + c }& \mathbf{=}& \mathbf{-5} \\ (2) & \mathbf {-a + b - c }& \mathbf{=}& \mathbf{15} \\ \hline (1)+(2): & 2b &=& -5+15 \\ & 2b &=& 10 \\ & \mathbf {b }& \mathbf{=}& \mathbf{5} \\ \hline (1) : & a+b+c &=& -5 \quad | \quad b=5 \\ & a+5+c &=& -5 \\ & \mathbf {a+c} &\mathbf {=}& \mathbf {-10} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline f(2) = 12: & f(2) = 12&=& a\cdot (2)^3+ b\cdot (2)^2+ c\cdot (2) \\ & 12 &=& 8a + 4b +2c \quad | \quad b=5 \\ & 12 &=& 8a + 4\cdot 5 +2c \\ & 12 &=& 8a + 20 +2c \\ & 8a + 2c &=& 12-20 \\ & 8a + 2c &=& -8 \quad & | \quad :2 \\ & \mathbf {4a + c}& \mathbf{=}& \mathbf{-4}\qquad (4) \\ \hline \end{array}\)

\(\mathbf{(4)-(3):}\)

\(\begin{array}{|lrcll|} \hline (4) & \mathbf {4a + c}& \mathbf{=}& \mathbf{-4} \\ (3) & \mathbf {a+c} &\mathbf {=}& \mathbf {-10}\\ \hline (4)-(3): & 3a &=& -4+-(-10)\\ & 3a &=& 6 \\ & \mathbf {a }& \mathbf{=}& \mathbf{2} \\ \hline (3) : & a+c &=& -10 \quad | \quad a=2 \\ & 2+c &=& -10 \\ & \mathbf {c} &\mathbf {=}& \mathbf {-12} \\ \hline \end{array}\)

\(\mathbf{\text{The $x$-intercepts of the graph of $~\boxed{f(x)=2x^3+5x^2-12x}$:}}\)

\(\begin{array}{|rcll|} \hline 2x^3+5x^2-12x &=& 0 \\ x\cdot (2x^2+5x-12) &=& 0 \\ \hline \mathbf{x_1} & \mathbf{=}& \mathbf{ 0 } \\ \hline 2x^2+5x-12 &=& 0 \\ x &=& \dfrac{-5\pm \sqrt{25-4\cdot 2 \cdot (-12)} } {2\cdot 2} \\\\ x &=& \dfrac{-5\pm \sqrt{121} } {4} \\\\ x &=& \dfrac{-5\pm 11 } {4} \\\\ x_2 &=& \dfrac{-5+ 11 } {4} \\ \mathbf{x_2} & \mathbf{=}& \mathbf{ \dfrac32 } \\\\ x_3 &=& \dfrac{-5- 11 } {4} \\ \mathbf{x_3} & \mathbf{=}& \mathbf{ -4 } \\ \hline \end{array}\)

The x-intercepts of the graph of f(x) are \(0,\dfrac32, -4\)

heureka
Dec 7, 2018