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viva [34]
3 years ago
11

Which point is an x-intercept of the quadratic function f(x) = (x – 4)(x + 2)? (–4, 0) (–2, 0) (0, 2) (4, –2)

Mathematics
2 answers:
son4ous [18]3 years ago
5 0
<span>f(x) = (x – 4)(x + 2),
x-intercept means that f(x) = 0.

0 = (x-4)(x+2)
(x-4)=0, x= 4, point (4,0)
(x+2)=0, x = -2, point (-2,0)

This graph has 2 x-intercepts:  (4,0) and (-2,0).

From given answers we can choose only </span>(-2,0).
mel-nik [20]3 years ago
3 0

Answer:

The answer is (–2, 0)

I just took the quiz

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Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2
Verdich [7]

Answer:

(a) f'(x)=-\frac{2}{x^3}

(b) y=-0.25x+0.75

Step-by-step explanation:

The given function is

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

According to the first principle of the derivative,

f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}

Cancel out common factors.

f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}

By applying limit, we get

f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}

f'(x)=\frac{-2x)}{x^4}

f'(x)=\frac{-2)}{x^3}                         .... (2)

Therefore f'(x)=-\frac{2}{x^3}.

(b)

Put x=2, to find the y-coordinate of point of tangency.

f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}

Substitute x=2 in equation 2.

f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

y-y_1=m(x-x_1)

y-0.25=-0.25(x-2)

y-0.25=-0.25x+0.5

y=-0.25x+0.5+0.25

y=-0.25x+0.75

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0
3 years ago
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