What is the x intercept of the line?

Mathematics

1 answer:

Firlakuza [10]3 years ago

5 0

Answer:

(22, 0)

Step-by-step explanation:

The x-intercept is the point on the line where y=0.

Here, we can see that as y decreases by 20 from 40 to 20, x increases by 30 from -38 to -8. At these rates, if y continues to decrease by 20 from 20 to 0, x will increase by 30 from -8 to +22. That is, the point (22, 0) is a point on this line. These are the coordinates of the x-intercept.

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Alternate solution

You can go to the trouble to find a number that is the slope of the line, write the equation for the line, and solve that equation for the x-intercept. Using the numbers above, we find the slope to be -20/+30 = -2/3. Using any point, we can write the point-slope equation as ...

y -20 = -2/3(x +8) . . . . . . . . I like to choose the smallest numbers

Now, solving for y=0, we get

-20 = -2/3(x +8)

30 = x +8 . . . . . . . multiply by -3/2

22 = x . . . . . . . . . . subtract 8

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Comment on solution methods

I am always looking for the easiest, quickest way to solve a problem, preferably without pencil and paper. The first step in solving any problem is to look at what you are given, and at what you are asked to find. (I had a college professor that required homework to be structured in 3 sections: Given, Find, Solution. It is helpful to think that way.)

Here, we're given a table of x- and y-values and asked to find the table entry (x=??) for y=0. The given y-values decrease by 10, but there is a pair of them that decrease by the same amount as the difference from 20 to 0. That is, the sequence 40, 20, 0 is an arithmetic sequence. Then the corresponding x-values of -38, -8, ?? will likewise be an arithmetic sequence. All we need to do is find the difference of numbers in that sequence and use it to find the value of ??. The difference -8 -(-38) = 30, so the difference ??-(-8) = 30. ?? = 22.

Our solution takes one subtraction and one addition, both of which we can do in our head.

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One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
$(x-h)^2+(y-k)^2=r^2$
the center is (h,k) and the radius is r

and the distance formula is
distance between $(x_1,y_1)$ and $(x_2,y_2)$ is
$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle

so
$(x+2)^2+(y-3)^2=18$
$(x-(-2))^2+(y-3)^2=(\sqrt{18})^2$
$(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2$

the radius is $3\sqrt{2}$
center is (-2,3)

find distance between (8,4) and (-2,3)

$D=\sqrt{(8-(-2))^2+(4-3)^2}$
$D=\sqrt{(8+2)^2+(1)^2}$
$D=\sqrt{10^2+1}$
$D=\sqrt{100+1}$
$D=\sqrt{101}$

$r=3\sqrt{2}$ ≈4.2
$D=\sqrt{101}$ ≈10.04

do r<D

(8,4) is outside the circle

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