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How to Find The X and Y Intercepts

 

The x-intercepts are points where the graph of a function or an equation crosses or "touches" the x-axis of the Cartesian Plane. You may think of this as a point with y-value of zero.
  • To find the x-intercepts of an equation, let y = 0 then solve for x.
  • In point notation, it is written as ( x , 0 )
x-intercept of a line x-intercepts of a parabola
shows the x intercept of a straight line shows the x intercept of a parabola
The y-intercepts are points where the graph of a function or an equation crosses or "touches" the y-axis of the Cartesian Plane. You may think of this as a point with x-value of zero.
  • To find the y-intercepts of an equation, let x = 0 then solve for y.
  • In point notation, it is written as ( 0 , y )
y-intercept of a line y-intercept of a parabola
   
diagram that shows the y intercept of a line diagram that shows the y intercept of a quadratic function

 

 

 


 

Example 1: From the graph, describe the x and y intercepts using point notation.

the graph of the quadratic function f(x)=-x^2-2x+3

 

The graph crosses the x-axis at x=1 and x=-3, therefore, we can write the x-intercepts as points (1,0) and (-3,0).

Similarly, the graph crosses the y-axis at y=3. Its y-intercept can be written as the point (0,3).

 


 

Example 2: Find the x and y intercepts of the line y=-2x+4.

 

To find the x-intercepts algebraically, we let y = 0 in the equation and then solve for values of x. In the same manner, to find for y-intercepts algebraically, we let x = 0 in the equation and then solve for y.

x-intercepts

( let y = 0, then solve for x )

y-intercepts

( let x = 0, then solve for y)

solving the equation y=-2x+4,  if y=0 then x=2 solving the equation y=-2x+4,  if x=0 then y=4
written as point: written as point: (0,4)

Here's the graph to verify that our answers are correct.

graph of the equation y=-2x+4 showing x=2 (x-intercept) and y=4 (y-intercept)(2,0)

 


 

Example 3: Find the x and y intercepts of the quadratic equation y=x^2-2x-3.

The graph of this quadratic equation is a parabola. We expect it to have a "U" shape where it would either open up or down.

To solve for the x-intercept of this problem, you will factor a simple trinomial. Then you set each binomial factor equal to zero and solve for x.

 

x-intercepts

(let y = 0, then solve for x)

y-intercepts

(let x = 0, then solve for y)

x1 = -1 and x2=3 y=-3
as points: (-1,0) and (3,0) as point: (0,-3)

 

Our solved values for both x and y intercepts match with the graphical solution.

 

graph of the equation y=x^2-2x-3 in xy axis showing where the x and y intercepts are located

 


 

Example 4: Find the x and y intercepts of the quadratic equation y=3x^2+1.

This is an example where the graph of the equation has a y-intercept but without an x-intercept.

  • Let's find the y-intercept first because it's extremely easy! Plug in x = 0 then solve for y.
showing the solution to obtain the y-intercept of  y=3x^2+1 which is y=1

 

  • Now for the x-intercept. Plug in y = 0, and solve for x.
x = no solution

The square root of a negative number is imaginary. This suggests that this equation does not have an x-intercept!

The graph can verify what's going on. Notice that the graph crossed the y-axis at (0,1); but never did with the x-axis.

graph of a parbola with y intercept at (0,1)  but no x-intercept

 


 

Example 5: Find the x and y intercepts of the circle (x+4)^2+(y+2)^2=8.

This is a good example to illustrate that it is possible for the graph of an equation to have x-intercepts but without y-intercepts.

 

x-intercepts

(let y = 0, then solve for x)

y-intercepts

(let x = 0, then solve for y)

x1=-2 and x=-6 y= no solution
as points: (-2,0) and (-6,0) none

 

When solving for y, we arrived at the situation of trying to get the square root of a negative number. The answer is imaginary, thus, no solution. That means, the equation doesn't have any y-intercepts.

The graph verifies that we are right for the values of our x-intercepts, and it has no y-intercepts.

 

graph of the equation of circle  (x+4)^2+(y+2)^2=8 showing its x-intercepts at (-6,0) and (-2,0)

 

 
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