All categories

3 years ago

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

2 answers:

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

You might be interested in

Please helpppp

Annette [7]

Answer:

1. 41

2. 7.857 or 8

3. 21

3 0

3 years ago

Read 2 more answers

Add, subtract, multiply, or divide the following fractions. Remember to find the LCD first.

Nat2105 [25]

lcd 210

1/5=42/210

1/7=30/310

3/21=30/210

42+30+30=102/210

reduced 51/105

4 0

3 years ago

The information given represents lengths of sides of a right triangle with c the hypotenuse. Find the correct missing length to

klasskru [66]

A² = c² - b²
a² = 21² - 13 ²
a² = 441 - 169
a² = 272
a = √272 = 16.49242 ≈ 16.49
Answer : D ) 16.49

5 0

3 years ago

Read 2 more answers

Find x if the area of the rectangle below is 124 square inches. WHAT IS THE

Tomtit [17]

Answer:

Step-by-step explanation:

(6x+1)4 = 124

6x+1 = 31

x = 5

4 0

3 years ago

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