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Novay_Z [31]
3 years ago
9

Find a parabola with equation y = ax2 + bx + c that has slope 5 at x = 1, slope −11 at x = −1, and passes through the point (2,

18).
Mathematics
1 answer:
azamat3 years ago
6 0

By "slope" I assume you mean slope of the tangent line to the parabola.

For any given value of <em>x</em>, the slope of the tangent to the parabola is equal to the derivative of <em>y</em> :

y=ax^2+bx+c\implies y'=2ax+b

The slope at <em>x</em> = 1 is 5:

2a+b=5

The slope at <em>x</em> = -1 is -11:

-2a+b=-11

We can already solve for <em>a</em> and <em>b</em> :

\begin{cases}2a+b=5\\-2a+b=-11\end{cases}\implies 2b=-6\implies b=-3

2a-3=5\implies 2a=8\implies a=4

Finally, the parabola passes through the point (2, 18); that is, the quadratic takes on a value of 18 when <em>x</em> = 2:

4a+2b+c=18\implies2(2a+b)+c=10+c=18\implies c=8

So the parabola has equation

\boxed{y=4x^2-3x+8}

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Answer:

c.

Step-by-step explanation:

Let x be the number of adults in Tim's family. Then (17 - x) is the number of children in Tim's family.

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2nd Semester Unit 1 Extra Credit!!! Can someone help me please?
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