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3 years ago

10

Given point P (3, 4). What is the distance of point P from (a) x axis (b) y axis?

1 answer:

vova2212 [387]3 years ago

6 0

Answer:

a) 3

b) 4

Step-by-step explanation:

The point P is(3,4)

which means that 3 is the point P's x- coordinate and 4 is it's y-coordinate

Because we use the axes as our reference point when plotting points,

the distance of P from x-axis and y-axis would be the point itself.

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2 years ago

Which graph represents the function of f(x) = the quantity of 9 x squared plus 9 x minus 18, all over 3 x plus 6

lora16 [44]

Answer:

The answer is the option C

graph of $3x$ minus $3$ , with discontinuity at negative $2$ , negative $9$

Step-by-step explanation:

we have

$f(x)=\frac{9x^{2}+9x-18}{3x+6}$

Simplify

$f(x)=9\frac{(x^{2}+x-2)}{3(x+2)}$

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

Step 1

Convert to a factored form the numerator

$x^{2}+x-2=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+x=2$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+x+0.25=2+0.25$

$x^{2}+x+0.25=2.25$

Rewrite as perfect squares

$(x+0.5)^{2}=2.25$

Square root both sides

$x+0.5=(+/-)1.5$

$x=-0.5(+/-)1.5$

$x=-0.5+1.5=1$

$x=-0.5-1.5=-2$

so

$x^{2}+x-2=(x-1)(x+2)$

Step 2

Simplify the function f(x)

$f(x)=3\frac{(x^{2}+x-2)}{(x+2)}=3\frac{(x-1)(x+2)}{(x+2)}$

The domain of the function f(x) is all real numbers except the number $x=-2$

Because the denominator can not be zero

$f(x)=3\frac{(x-1)(x+2)}{(x+2)}=3(x-1)=3x-3$

$f(x)=3x-3$ ------> with a discontinuity at $x=-2$

$f(-2)=3(-2)-3=-9$

The discontinuity is at point $(-2,-9)$

the answer in the attached figure

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