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Q: How do you find the x and y intercepts of an equation?

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Given the linear equation 3x - 2y^6 = 0, the x and y intercepts are found by replacing the x and y with 0. This gives the intercepts of x and y where both = 0.

Slope and y-interceptIn the general equation of the line, y=mx+b, the slope is m and the y-intercept is b. You can find the y-intercept for a linear equation in any form by setting x to zero and solving for y. Similarly, you can find the x-intercept by setting y to zero and solving for x.In the above general equation, the x-intercept is -b/m.

Assuming the equation is 3x + y = 15, then the x-intercept is (5, 0) and the y-intercept is (0, 15).

There are no intercepts because the curve, xy = 4 is asymptotic. When x = 0 (where the y intercept would be) y is infinite, and conversely, when y = 0 x is infinite.

Gradients can be worked out by: 1. gradient formula, suppose the two points are (x1,y1); (x2,y2) then the gradient=(y2-y1)/(x2-x1) 2. rise/run Intercepts can be found by: 1. to find the x-intercept substitute y=0 into the equation of the line 2. to find the y-intercept substitute x=0 into the equation of the line

Related questions

Given the linear equation 3x - 2y^6 = 0, the x and y intercepts are found by replacing the x and y with 0. This gives the intercepts of x and y where both = 0.

I believe that you need an equation to solve for the x and y intercepts.

The 'x' and 'y' intercepts of that equation are both at the origin.

If there is no y, then the equation is of the form x = c where c is some constant value. And so the line intercepts the x axis at (c,0).

In the equation y = f(x), Put x = 0 and solve for y. Those are the y intercepts. Put y = 0 and solve for x. Those are the x intercepts.

If "a" is negative then the graph is a cap. Find the x intercepts. Average the two x intercepts and substitute that into the equation it will give you the y.

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The equation 9=3y has the x-intercept (0,0) and the y-intercept (0,3).

A circle represented by an equation x^2 + y^2 = r^2 or a circular object represented by an equation Ax^2 + By^2 = r^2 has 2 y-intercepts and 2 x-intercepts.

The y-intercept is c in the standard form. The x-intercept is -c/m.

x2 + y - 49 = 0At the x-intercepts, y=0:x2 - 49 = 0(x+7)(x-7) = 0x = -7 and x = +7At the y-intercept, x=0:y - 49 = 0y = 49General Solutionset x=0 to find your y-intercept and set y=0 to find your x-intercept. That's how you will always find your intercepts, no matter what the equation is.Ex. So if you find that when x=0, y=a, then the y-intercept is at the point (0,a).Similarly, if you find that when y=0, x=b, then the x-intercept is at the point (b,0).You can solve any problem from here.

From the equation, the y intercept is simply determined by setting x = 0. The x intercept(s) are generally much harder to find: you will need to find the solutions of y = 0 [or f(x) = 0]. From the graph the intercepts are the coordinates of the points at which the graph crosses the axes.

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