The question does not contain an equation (or inequality) but an expression. An expression cannot have intercepts.
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.
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.
I believe that you need an equation to solve for the x and y intercepts.
If you mean y = x+4 and y = x-2 then the slopes would be parallel but with different y intercepts
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The 'x' and 'y' intercepts of that equation are both at the origin.
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.
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.
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 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).
I believe that you need an equation to solve for the x and y intercepts.
Yes it can. A linear equation in the form of y=mx+b can always be graphed used the x and y intercepts.
To find the intercepts of the equation (y = x^4 - 2x^2 - 8), we need to determine where the graph intersects the x-axis and y-axis. For the y-intercept, set (x = 0), yielding (y = -8), so the y-intercept is (0, -8). To find the x-intercepts, set (y = 0) and solve the equation (x^4 - 2x^2 - 8 = 0); this can be factored or solved using substitution methods, leading to the x-intercepts at approximately (x \approx 2.414) and (x \approx -2.414).
The x-intercepts are obtained by solving the equation for those value of x for which y or f(x) = 0 : where f(x) is the equation of the curve or line. If f(x) is a straight line, and the equation is in the form y = mx + c, then y = 0 gives x = -c/m For a quadratic, of the form y = ax2 + bx + c, the x-intercents are the root sof the equation, ie [-b ± sqrt(b2 - 4ac)]/(2a). The intercepts are real only when the discriminant, b2 - 4ac is non-negative.
If you mean y = x+4 and y = x-2 then the slopes would be parallel but with different y intercepts
No; it means draw the curve.
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