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2 years ago

Let f(x) = 3x2 - 6x + 2. Find f(-2). A) 20 B) 22 C) 24 D) 26

Mathematics

2 answers:

Nadusha1986 [10]2 years ago

8 0

Answer:

D) 26

Step-by-step explanation:

f(x) = 3x^2 - 6x + 2

Let x = -2

f(-2) = 3(-2)^2 - 6(-2) + 2

=3(4) +12+2

= 12 +12+2

=26

FrozenT [24]2 years ago

8 0

For this case we have a function of the form $y = f (x).$

Where:

$f (x) = 3x ^ 2-6x + 2.$

We must find the value of the function when $x = -2.$

Substituting we have:

$f (-2) = 3 (-2) ^ 2-6 (-2) +2\\f (-2) = 3 * 4 + 12 + 2\\f (-2) = 12 + 12 + 2\\f (-2) = 26$

Thus, the value of the function is 26.

Answer:

Option D

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The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

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Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$