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2 years ago

The x-intercepts of a parabola are (-5, 0) and (3, 0). What is the function of the parabola?

Mathematics

1 answer:

weeeeeb [17]2 years ago

3 0

The function of the parabola will be f(x) = -2(x²+2x-15)

<h3>Equation of a parabola</h3>

The equation of a parabola are quadratic in nature.

If s graph has intercepts (a, 0) and (b, 0), the function will be equivalent to;

f(x) = (x-a)(x-b)

Since we are given the x-intercepts (-5, 0) and (3, 0), the function of the parabola will be:

f(x) = (x+5)(x-3)

Expand

f(x) = x^2 - 3x + 5x - 15

f(x) = x^2 + 2x - 15

Hence the function of the parabola will be f(x) = -2(x²+2x-15)

Learn more on function of a parabola here: brainly.com/question/4061870

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2 years ago

Question 1- Whats the derivative of: A) f(x)= 4 cos(x) + ln(x+1)<br> B) f(x)= sec(x) X tg(x)

Kryger [21]

The derivatives for this problem are given as follows:

a) $f^{\prime}(x) = -4\sin{x} + \frac{1}{x + 1}$

b) $f^{\prime}(x) = \sec{x}\tan^{2}{x} + \sec^3{x}$ .

<h3>What is the derivative of the sum?</h3>

The derivative of the <u>sum is the sum of the derivatives</u>.

In this problem, the function is:

$f(x) = 4\cos{x} + \ln{(x + 1)}$

Using a derivative table for the derivatives of the cosine and the ln, the derivative of the function is:

$f^{\prime}(x) = (4\cos{x})^{\prime} + (\ln{(x + 1)})^{\prime}$

$f^{\prime}(x) = -4\sin{x} + \frac{1}{x + 1}$

What is the product rule?

The derivative of the product is given as follows:

$(f(x) \times g(x))^{\prime} = f^{\prime}(x)g(x) + g^{\prime}(x)f(x)$

In this problem, we have that:

$f(x) = \sec{x}, f^{\prime}(x) = \sec{x}\tan{x}$ .

$g(x) = \tan{x}, f^{\prime}(x) = \sec^2{x}$ .

Hence the derivative is:

$f^{\prime}(x) = \sec{x}\tan^{2}{x} + \sec^3{x}$ .

More can be learned about derivatives at brainly.com/question/2256078

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