Help? Very confused on question 8.

Mathematics

1 answer:

Fynjy0 [20]3 years ago

4 0

For part a: you just need to find how far the vertex has been moved from the origin, or the point (0,0). As the vertex is at the point (2,-3), it has been translated right 2 horizontally and down 3 vertically.

For part b: you use the info found in part a to create the equation in the form of y=A(x-h)^2+k. In this case, A =1, so you can ignore it. The h value is the horizontal distance the vertex has been moved. Since it has been moved right 2, this part of the equation would be (x-2). I know it seems like it should be plus 2, but values in parentheses come out opposite. For the k value, find the vertical shift, which is down3, or -3.
Now that you have h and k, substitute them back into the equation.
Your final answer for part b is: y=(x-2)^2 -3.

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Consider the integral 8 (x2+1) dx 0 (a) Estimate the area under the curve using a left-hand sum with n = 4. 250 Is this sum an o

Leya [2.2K]

Answer:

(a) 120 square units (underestimate)

(b) 248 square units

Step-by-step explanation:

See the attachment for a diagram of the areas being summed (in orange). This is the sum of the first 4 table values for f(x), each multiplied by 2 (the width of the rectangle). Quite clearly, the curve is above the rectangle for the entire interval, so the rectangle area underestimates the area under the curve.

left sum = 2(1 + 5 + 17 + 37) = 2(60) = 120 . . . . square units

<u>(b) right sum</u>

The right sum is the sum of the last 4 table values for f(x), each multiplied by 2 (the width of the rectangle). This sum is ...

right sum = 2(5 +17 + 37 +65) = 2(124) = 248 . . . . square units