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3 years ago

find the equation of a cubic function whose graph passes through points (3,0) and (1,4) and is tangent to x-axis at the origin

Mathematics

1 answer:

Tomtit [17]3 years ago

3 0

Answer:

y = -2x^2(x - 3)

Step-by-step explanation:

Preliminary Remark

If a cubic is tangent to the x axis at 0,0

Then the equation must be related to y = a*x^2(x - h)

(3,0)

If the cubic goes through the point (3,0), then the equation will become

0 = a*3^2(3 - h)

0 = 9a (3 - h)

0 = 27a - 9ah

from which h = 3

From the second point, we get

4 = ax^2(x - 3)

4 = a(1)^2(1 - 3)

4 = a(-2)

a = 4 / - 2

a = -2

Answer

y = -2x^2(x - 3)

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3 years ago

You have the set of all the integers from (-16) to 26, inclusive. You take two of the

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2 years ago

Read 2 more answers

B. Determine the values of a and b so that f is continuous. Please help!

wlad13 [49]

Answer:

following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

$\lim_{x\to 2}+f(x) = \lim_{x\to 2}+ (ax^2-bx+3) = 4a-2b+3\\\\ \to 4a-2b+3 = 4\\\\ \therefore \ \ 4a-2b = 1\\\\$

The one-sided limits of F(x) at x = 3 must be equivalent for f(x) to be continuous at x = 3.

$\to \lim_{x\to 3}- f(x) = \lim_{x\to 3}- (ax^2-bx+3) = 9a-3b+3\\\\ \to \lim_{x\to 3}+ f(x) = \lim_{x\to 3}+ (2x-a+b) = 6-a+b\\\\$

So,

$\to 9a-3b+3 = 6-a+b\\\\\therefore\\ \to 10a -4b = 3$

$\to 4a - 2b = 1......(a)\\\to 10a - 4b = 3.......(b)\\$

In equation a multiply the by -2 and then add in the equation b:

$\to -8a + 4b = -2\\\to 10a - 4b = 3\\ \to 2a = 1\\\\ \to a = \frac{1}{2}\\\\ \to \ 4(\frac{1}{2}) - 2b = 1\\\\ \to 2 - 2b = 1\\\\ \to -2b = -1\\\\ \to b = \frac{1}{2}$

So, the value of $a \ and \ b= \frac{1}{2}$

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