The y intercept of any parabola can be obtained by substituting the value of x=0 wich is the eqaution of y axis. FYI three cases will be possible.1.THE PARABOLA TOUCHES THE Y AXIS IN 2 POINTSHere when you substitute x=0,you will get a quadratic equation in y having 2 real values.2.THE PARABOLA TOUCHES THE Y AXIS IN 1 POINTHere you will get one and reapeating values of y.3.THE PARABOLA DOES NOT TOUCH THE Y AXIS.Here the quadratic eqation roots will be imaginary.
Calculate the slope, y intercept, and x intercept for the following equation: y= 2+ x / 5?
if you are given an equation that looks like y=mx + b.. the m represents the slope, which is rise/run... and the b, represents the value of y, when x is zero.. so to use, just let x=0, then see what b is.. for example, y=(1/3)x -6 the slope means rise one, run 3..rise means up, run means always to the right... if you get a negative the rise will go downwards.. the x intercept is the opposite of the y intercept. in other words, let the y=0 and solve for the x value.. in the example above, you would set 0=(1/3)x -6 means, that (1/3)x=6 or x=18 you would find the intercept at (18,0) for the x intercept, and the y intercept would be,.... you guess it.. (0, -6). hope that helps..
The y-intercept of a line is the y-coordinate of the point of intersection, (x, y), of the line and the y-axis of a rectangular coordinate system. Since every point on the y-axis has an x-coordinate of zero, i.e., (0,y), then in order to find the y-intercept of a given line, we must first set x = 0 in the equation algebraically describing the line and then solve for y as follows:y = 5x - 8y = 5(0) - 8y = 0 - 8y = -8Therefore, the y-intercept of the line described by the given equation is -8, and the point of intersection, (x, y), between the line and the y-axis is: (0, -8).
For the x intercept, set y = 0 and solve for x.y = 2x + 70 = 2x + 72x = -7x = -7/2 = -3.5So the x intercept is at ( -3.5 , 0 )For the y intercept set x = 0 and solve for y.y = 2(0) +7y = 7So the y intercept is ( 0, 7 )Go to Desmos, enter the equation y = 2x + 7 and look at the graph of this equation.
I need help finding X and Y intercepts of the equation(s)?
To find the x-intercept, plug in 0 for y, and to find the y-intercept, plug in 0 for x. 1.) 2x-5y=20 2(0)-5y=20 -5y=20 y=-4 <===y-intercept 2x-5(0)=20 2x=20 x=10 <===x-intercept -------------------------- 2.) x+5y=8 (0)+5y=8 5y=8 y=8/5 <===y-intercept x+5(0)=8 x+0=8 x=8 <===x-intercept -------------------------- 3.) 3x-y=-7 3(0)-y=-7 -y=-7 y=7 <===y-intercept 3x-(0)=-7 3x=-7 x=-7/3 <===x-intercept
X- and y- intercepts help?
To find the x and y intercepts, all you have to do is: If you're looking for the x-intercept, set y = 0 and solve! If you're looking for the y-intercept, set x = 0 and solve. But by looking at the polynomial, you can determine the roots, meaning, where the function crosses the x-axis. f(x) = x (x - 2) (x - 4) x = 0, +2, and +4 Also, the function always crosses the x-axis, UNLESS it is raised to an exponent. For example, if it is (x - 4)^2, then it will curve the other way at x = 4, and not cross it (double root). and if it is (x - 4)^3, it will cross it but with an inflection point.
Calculus help! Tangent line and x/y intercepts, derivatives, and velocity!?
Hope this helps, 1) Find the derivative of the function (The slope of the tangent line), then substitute in the point given to find the slope of the specific tangent line that touches the curve at the given point. y=x^3, y'=3x^2, y'(-4)=48. Therefor the slope, m, of the tangent line at (-4,-64) is 48. Now find the equation of the tangent line, which is in the form y = mx +b: y = 48x + b Solve for b, using the given point (-4, -64): -64 = 48(-4) + b, -64 = -192 + b, b = 128; y = 48x +128; The x-intercept occurs when y=0, then solve for x. 0 = 48x +128, 48x = -128. x = -128/48 = -8/3 (-8/3, 0) The y-intercept occurs when x=0, then solve for y. y = 0 + 128, y = 128 (0,128) That's it =) 2) d/dx (3u-2v+2uv) = 3u' - 2v' + 2(u'v + v'u) *You have to use the product rule for 2uv That is the general solution, so now substitue in for x=2 = 3u'(2) - 2v'(2) + 2[u'(2)v(2) + v'(2)u(2)] = 3(-9) - 2(2) + 2[(-9)(1) + (2)(3)] = -27 - 4 +2(-3) = -31 - 6 = -37 3) Displacement is total movement (change in position). So to find the position you find how far it moved. So the question is what is the difference between the final position and initial position. In math terms this is: s(5) - s(0) = 8 - 3 = 5 If you have a function for position, if you take the derivative, the gives you a function for the velocity at any point. So find the derivative of s(t), s'(t) = v(t). Then substitute in t=4 to give the instantaneous velocity at that point. s'(t) = v(t) = 2t - 4; v(4) = 8 - 4 = 4 Feel free to give a follow up question and I can explain specific things further if you would like. Mike