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2 years ago

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What value of x makes the

2 answers:

loris [4]2 years ago

6 0

Answer:

$x = -12$

Step-by-step explanation:

<u>Step 1: Add 1/4x to both sides</u>

$-\frac{1}{4}x - 12 = x + 3$

$-12 - \frac{1}{4}x + \frac{1}{4}x = x + 3 + \frac{1}{4}x$

$-12 = \frac{5}{4}x + 3$

<u>Step 2: Subtract 3 from both sides</u>

$-12 - 3 = \frac{5}{4}x + 3 - 3$

$-15 = \frac{5}{4}x$

<u>Step 3: Multiply both sides by 4/5</u>

$-15*\frac{4}{5}=\frac{5}{4}x * \frac{4}{5}$

$-12 = x$

Answer: $x = -12$

Vlada [557]2 years ago

6 0

Answer: x = -18.75

-1/4x - 12 = x + 3

Add 12x from both sides to get
-1/4x = x + 15

Then minus x from both sides
-5/4x = 15

The divide by -5/4 to get
x = -18.75

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F(x) = x^2+x-2/x^2-3x-4

Andrej [43]

i. Domain and Range

The given function is

$f(x)=\frac{x^2+x-2}{x^2-3x-4}$

The domain of this function is,

$x^2-3x-4\ne 0$

$(x-4)(x+1)\ne 0$

$x\ne4,xne -1$

The range refers to the y-values for which x is defined. x is defined for all values of y.

The range is all real numbers. See graph

ii. x-and-y-intercept

For x- intercept intercept we put $f(x)=0$

This implies that;

$\frac{x^2+x-2}{x^2-3x-4}=0$

This will give us

$x^2+x-2=0$

$\Rightarrow x^2+x-2=0$

$\Rightarrow x^2+2x--x-2=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow (x+2)(x-1)=0$

$\Rightarrow (x+2)=0,(x-1)=0$

$\Rightarrow x=-2,x=1$

The x-intercepts are $(-2,0),(1,0)$

For y-intercept, we put

$x=0$ to obtain;

$f(0)=\frac{0^2+0-2}{0^2-3(0)-4}$

$f(0)=\frac{1}{2}$

The y-intercept is

$(0,\frac{1}{2})$

iii. Horizontal asyptote

Since degree of the numerator and the denominator are the same, there is a horizontal asymptote

To find the horizontal asymptote.

We divide the leading coefficient of the numerator by the leading coefficient of the denominator.

The horizontal asymptote is $y=\frac{1}{1}=1$

iv. Vertical asymptote

To find the vertical asymptote, we equate the denominator to zero to get;

$x^2-3x-4=0$

This implies that;

$x^2+x-4x-4=0$

Split the middle term

$x(x+1)-4(x+1)=0$

Factor

$(x+1)(x-4)=0$

Factor further

$(x+1)=0,(x-4)=0$

$x=-1,x=4$

The vertical asymptotes are $x=-1,x=4$

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