+0

# help

0
70
2
+262

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?

Apr 10, 2019

#1
+5172
+3

$$f(x) = a x^3 + b x^2 + c x + d\\ f(-1) = -a + b - c + d = 15\\ f(0) = d = 0\\ f(1) = a + b + c + d = -5\\ f(2) = 8a+4b+2c+d=12\\ \text{I like to use matrices, there are other ways to solve the following}\\ \begin{pmatrix} -1&1 &-1 \\ 1 &1 &1 \\ 8 &4 &2 \end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix}= \begin{pmatrix}15\\-5\\12\end{pmatrix}$$

$$\text{Gaussian reducing this we get}\\ \begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}2\\5\\-12\end{pmatrix}$$

$$f(x) = 2x^3 + 5x^2 -12x = \\ x(2x^2 + 5x - 12) = \\ x(2x-3 )(x+4 )\\ \text{x intercepts at }x \in \left\{0,~\dfrac 3 2,~-4\right\}$$

.
Apr 10, 2019
#2
+262
0

thanks!

Apr 10, 2019