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2 years ago

What is the y intercept of the graph of g(x)=(x+1)(x-2)(x+5)

Mathematics

1 answer:

Ymorist [56]2 years ago

4 0

Answer:

The y intercept is -10

Step-by-step explanation:

Given

$g(x)=(x+1)(x-2)(x+5)$

Required

The y intercept

To calculate the y intercept, we set $x = 0$

So, we have:

$g(x)=(x+1)(x-2)(x+5)$

$g(0)=(0+1)(0-2)(0+5)$

$g(0)=(1)(-2)(5)$

$g(0)=-10$

You might be interested in

If f(x)=3x-2 find f-1 (13)

SOVA2 [1]

Answer:

Firstly

f(x)=3x-2=y then

Interchanging x and y we get

x=3y-2

x+2=3y

y=(x+2)/3

f-1(x)=(x+2)/3

f-1(13)=(13+2)/3

f-1(13)=5

3 0

3 years ago

Can anyone help me please

Nady [450]

Yo sup??

since WY is the perpendicular bisector we can say

3x-5=2x+3

x=8

Hope this helps

6 0

2 years ago

Help me please ;-;!!!!! Use a coordinate grid to create a map of a town with at least five different locations, such as a house,

Sholpan [36]

A. Our diagram have six location: my house (2,1), a movie theater (5,3), the school (6,6), the police station (3,-2), the airport (-3,3), and the mall (-2,-3)

B. Distance between my house and the movie theater:
To calculate the distance we are going to use the points (2,1) and (5,3) to create a right triangle with tow legs of measures 3 and 2; the hypotenuse of the right triangle will be the distance between my house and the movie theater.
Using the Pythagorean theorem:
$d^2=3^2+2^2$
$d^2=9+4$
$d^2=13$
$d= \sqrt{13}$
We can conclude that the distance between my house and the movie theater is $\sqrt{13}$

Our second distance is the distance between the police station and the airport:
This time we are using the points (3,-2) and (-3,3) to create a right triangle with legs of measures 6 and 5; the hypotenuse of our triangle will be the the distance between our points:
$d^2=6^2+5^2$
$d^2=36+25$
$d^2=61$
$d= \sqrt{61}$
We can conclude that the distance between the police station and the airport is $\sqrt{61}$

C. Distance form the park to the Fire Station:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance from the park to the Fire Station is $\sqrt{10}$

Distance from the stadium to the mall:
$d^2=2^2+1^2$
$d^2=4+1$
$d^2=5$
$d= \sqrt{5}$
We can conclude that the distance from the stadium to the mall is $\sqrt{5}$

D. What is the distance between your house and the mall?
Answer:
$d^2=4^2+4^2$
$d^2=16+16$
$d^2=32$
$d= \sqrt{32}$
$d=4 \sqrt{2}$
We can conclude that the distance between my house and the mall is $4 \sqrt{2}$

What is the distance between the movie theater and the school?
answer:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance between the movie theater and the school is $\sqrt{10}$