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irga5000 [103]
1 year ago
6

The first steps in writing f(x) = 3x2 – 24x 10 in vertex form are shown. f(x) = 3(x2 – 8x) 10 (startfraction negative 8 over 2 e

ndfraction) squared = 16 what is the function written in vertex form? f(x) = 3(x 4)2 – 6 f(x) = 3(x 4)2 – 38 f(x) = 3(x – 4)2 – 6 f(x) = 3(x – 4)2 – 38
Mathematics
2 answers:
amid [387]1 year ago
8 0

Answer:

The answer is D!

Step-by-step explanation:

fredd [130]1 year ago
6 0

The given quadratic equation 3x^2 – 24x + 10 can be written in vertex form that is f(x) = 3(x – 4)2 – 38.

<h3>What are quadratic equations?</h3>

A quadratic equation is an equation of degree 2 and the standard form of a quadratic equation ax^{2} +bx+c where x is the variable and a,b, c are real numbers.

Following the steps already shown to transform the given function into the vertex form,

f(x) = 3(x^{2}  - 8x + 16) + 10 - 3(16)\\f(x) = 3(x - 4)(x - 4) + 10 - 48\\f(x) = 3(x - 4)2 - 38

Therefore, the function written in vertex form is f(x) = 3(x – 4)2 – 38.

Learn more about quadratic equations;

brainly.com/question/2263981

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i) The given function is

f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}

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ii) For vertical asymptotes, we simplify the function to get;

f(x)=\frac{(2x+1)}{(x+4)^2}

The vertical asymptote occurs at

(x+4)^2=0

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iii) The roots are the x-intercepts of the reduced fraction.

Equate the numerator of the reduced fraction to zero.

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2x=-1

x=-\frac{1}{2}

iv) To find the y-intercept, we substitute x=0 into the reduced fraction.

f(0)=\frac{(2(0)+1)}{(0+4)^2}

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v) The horizontal asymptote is given by;

lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0

The horizontal asymptote is y=0.

vi) The function has a hole at x-5=0.

Thus at x=5.

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

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