Can you please help me with this question with the steps? 8a^3 -2a^2+ 12a-3

1 answer:

6 0

I'm assuming you want to simplify the expression since it's not equal to anything.. you can't solve for "a" unless it's equal to something.

factor

8a³ - 2a²+ 12a - 3

2a²(4a - 1) + 3(4a -1)

(4a - 1) is a like term that can be combined further.

(2a² + 3)(4a - 1)

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I need help with this problem

lubasha [3.4K]

at x = 1 we have one tangent line and at x = 5 we have just another tangent line.

$f(x)=3x^2-15x\implies \left. \cfrac{df}{dx}=6x-15 \right|_{x=1}\implies \stackrel{\stackrel{m}{\downarrow }}{-9}~\hfill \left. 6x-15\cfrac{}{} \right|_{x=5}\implies \stackrel{\stackrel{m}{\downarrow }}{15}$

so we have the slopes, but what about the coordinates?

well, for the first one we know x = 1 and we also know f(x), let's use f(1) to get "y", and likewise we'll do the for the second one.

$\stackrel{x=1}{f(1)}=3(1)^2-15(1)\implies f(1)=-12\qquad \qquad (\stackrel{x_1}{1}~~,~~\stackrel{y_1}{-12}) \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-12)}=\stackrel{m}{-9}(x-\stackrel{x_1}{1}) \\\\\\ y+12=-9x+9\implies y=-9x-3 \\\\[-0.35em] ~\dotfill$

$\stackrel{x=5}{f(5)}=3(5)^2-15(5)\implies f(5)=0\qquad (\stackrel{x_1}{5}~~,~~\stackrel{y_1}{0}) \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{15}(x-\stackrel{x_1}{5})\implies y=15x-75$