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Question
- find the average rate of change from x=3 to x=5 (unit 1.2)
f(x)= -12x² - 2x + 1
find the average rate of change from x=2 to x=4? (unit 1.2)
f(x)= -3x³ -x + 3
- a boy drops his teddy bear off a 70 foot building. the height of the bear over time can be modeled by the function below, where t is time in seconds and h is the height in feet. what is the average rate of change from 1 second to 3 seconds? (unit 1.2)
h(t)= -16t²+4t+70
a bratty kid spits off a 60 foot balcony with an initial speed of 13 feet per second. what is the average rate of change from 2 seconds to 4 seconds? where t is time in seconds and h is the height in feet. (unit 1.2)
h(t)= -16t²+13t+60
- describe the transformations to the following function: (unit 1.3)
y= -f(x+11)^2 - 7
point (-1, 2) lies on the graph of the function f(x)
suppose the function g represents a transformation of function f, g(x) = f(x) + 3 (unit 1.3)
write the new ordered pair after the transformation.
Step1: Recall the formula for average rate of change
The average rate of change of a function \(y = f(x)\) from \(x=a\) to \(x = b\) is given by \(\frac{f(b)-f(a)}{b - a}\).
Step2: Calculate \(f(3)\) and \(f(5)\) for \(f(x)=-12x^{2}-2x + 1\)
- For \(x = 3\):
\(f(3)=-12\times(3)^{2}-2\times3 + 1=-12\times9-6 + 1=-108-6 + 1=-113\)
- For \(x = 5\):
\(f(5)=-12\times(5)^{2}-2\times5 + 1=-12\times25-10 + 1=-300-10 + 1=-309\)
Step3: Apply the average - rate - of - change formula
\(\frac{f(5)-f(3)}{5 - 3}=\frac{-309-(-113)}{2}=\frac{-309 + 113}{2}=\frac{-196}{2}=-98\)
Step4: Calculate \(f(2)\) and \(f(4)\) for \(f(x)=-3x^{3}-x + 3\)
- For \(x = 2\):
\(f(2)=-3\times(2)^{3}-2 + 3=-3\times8-2 + 3=-24-2 + 3=-23\)
- For \(x = 4\):
\(f(4)=-3\times(4)^{3}-4 + 3=-3\times64-4 + 3=-192-4 + 3=-193\)
Step5: Apply the average - rate - of change formula
\(\frac{f(4)-f(2)}{4 - 2}=\frac{-193-(-23)}{2}=\frac{-193 + 23}{2}=\frac{-170}{2}=-85\)
Step6: Calculate \(h(1)\) and \(h(3)\) for \(h(t)=-16t^{2}+4t + 70\)
- For \(t = 1\):
\(h(1)=-16\times(1)^{2}+4\times1 + 70=-16 + 4+70=58\)
- For \(t = 3\):
\(h(3)=-16\times(3)^{2}+4\times3 + 70=-16\times9+12 + 70=-144+12 + 70=-62\)
Step7: Apply the average - rate - of change formula
\(\frac{h(3)-h(1)}{3 - 1}=\frac{-62 - 58}{2}=\frac{-120}{2}=-60\)
Step8: Calculate \(h(2)\) and \(h(4)\) for \(h(t)=-16t^{2}+13t + 60\)
- For \(t = 2\):
\(h(2)=-16\times(2)^{2}+13\times2 + 60=-16\times4+26 + 60=-64+26 + 60=22\)
- For \(t = 4\):
\(h(4)=-16\times(4)^{2}+13\times4 + 60=-16\times16+52 + 60=-256+52 + 60=-144\)
Step9: Apply the average - rate - of change formula
\(\frac{h(4)-h(2)}{4 - 2}=\frac{-144-22}{2}=\frac{-166}{2}=-83\)
Step10: Analyze the transformation \(y=-f(x + 11)^{2}-7\)
- Horizontal shift: The graph of \(y = f(x)\) is shifted left by \(11\) units (because of \(x+11\)).
- Reflection: The graph is reflected about the \(x\) - axis (because of the negative sign in front of \(f\)).
- Vertical stretch/squeeze: Since it's \(f(x + 11)^{2}\), if \(|a|>1\) (here \(a = 1\) for the squaring of the function value, but if we consider the general form \(y = af(x - h)+k\), the squaring is a non - linear transformation. In terms of basic transformations:
- Vertical shift: The graph is shifted down by \(7\) units (because of \(-7\)).
Step11: Find the new point for \(g(x)=f(x)+3\)
If \((x,y)\) lies on \(y = f(x)\), then for \(y = g(x)=f(x)+3\), when \(x=-1\), \(y=f(-1)+3\). Since \(f(-1) = 2\), then \(g(-1)=2 + 3=5\). The new ordered pair is \((-1,5)\)
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- The average rate of change of \(f(x)=-12x^{2}-2x + 1\) from \(x = 3\) to \(x = 5\) is \(-98\).
- The average rate of change of \(f(x)=-3x^{3}-x + 3\) from \(x = 2\) to \(x = 4\) is \(-85\).
- The average rate of change of \(h(t)=-16t^{2}+4t + 70\) from \(t = 1\) to \(t = 3\) is \(-60\).
- The average rate of change of \(h(t)=-16t^{2}+13t + 60\) from \(t = 2\) to \(t = 4\) is \(-83\).
- The transformation of \(y=-f(x + 11)^{2}-7\) is a left shift of \(11\) units, reflection about the \(x\) - axis, and a down shift of \(7\) units.
- The new ordered pair for \(g(x)=f(x)+3\) with the point \((-1,2)\) on \(f(x)\) is \((-1,5)\).