What is the equation that gives you a heart on the graph?
I actually like this equation:
Math help (transforming parabola's)?
What you need to do is compare the parabolas of these equations to the standard y=x^2 parabola. When given problems like these I recommend using a good graphing calculator like the TI-84 because you can put in multiple equations and when you graph them all of the graphs appear at once. A.) This equation is compressed by a factor of 3, it is opened up, and its vertex is at (0,-8) B.) This equation is is opened up and the only change is that its vertex is at (6,4) C.) This equation is compressed by a factor of 4, it is opened downward, and its vertex is at (-3,-7) Also, if you don't have a calculator you can figure this all out by looking at the equations. A number in front of the x shows whether it is compressed or expanded (a number greater than 1 will result in compression and less than 1 will be an expansion), adding or subtracting a number after the x^2 will shift the vertex up and down on the y-axis, and number with the x (like (x-6)^2) will shift the vertex left and right (however unlike with the y-axis it is the opposite or the sign so adding to the x will shift it to the left and subtracting will shift it to the right), and then the sign of the x determines which way it faces (so a positive x like in a and b makes it open up and a negative x like in c where it is multiplied by a -4 makes it open downward).
When graphing an equation of the form y=mx+c, how do you know which variables represent m, x and c?
To know m,x & c, first express the equation in the form of a linear eqution.Example - Let's say we have a linear equation: 2x + y = 4Convert it in the form: y = mx + cThe equation will become: y = -2x + 4 So, m = -2 (Slope/gradient of the line)and, c = 4 (Y intercept)As, s = ut + 1/2(at^2) is a quadratic equation, hence it cannot be expressed in the form of y = mx + c.But, it can be expressed in the form of the graph of a quadratic equation of the form y = ax^2 + bx + c.Where, a, b, c are constants.This graph can be represented by a parabola. Let's try to express the motion of equation into that form.s = (1/2a)t^2 + ut + 0here, a = 1/2a (Assuming acceleration is constant) b = u (initial speed is constant) c = 0 (Constant) y = s x = tNow, the parabola of a quadratic equation depends upon the nature of the constant 'a' and discriminant 'D'(which is b^2 - 4ac).Assume, a>0 (Positive acceleration)And, with c = 0, D will be b^2 > 0So, the graph will be a parabola opening upwards and would look like the one below.where,Y axis - SpeedX axis - TimeShaded region - DistanceTop-most curve - Equation of motion Hope this helps. Feel free to correct me if something is wrong.
How are graphs of the functions obtained from the graph of f?
A). You have a graph, y = -f(x) B). Start with the same graph as in A), except it's twice as steep (because of the 2) and it's shifted down one unit (because of the -1) c) Start with the same graph as in A), except it's shifted three units to the right (because of the x-3) and shifted up two units (because of the +2)
Describe how the graph if f varies as c varies.?
graph several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and minimum points and inflection points move when c changes. you should also identify any transitional values of c at which the basic shape of the curve changes. f(x)=(cx)/(1+c^(2)x^(2))
Why is [math]x^2 + y^2 = r^2[/math] the equation of a circle?
A conic section (or simply conic) is a curve obtained as the intersection of the surface of a cone with a plane. The three types of conic section are thehyperbola, the parabola, and the ellipse. The circle is a special case of the ellipse, and is of sufficient interest in its own right that it was sometimes called a fourth type of conic section.In the Cartesian coordinate system, the graph of a quadratic equation in two variables is always a conic section and all conic sections arise in this way. The most general equation is of the formThe conic sections described by this equation can be classified by the Discriminant of the equation:if [math]B2−4AC<0[/math], the equation represents an ellipse;if [math]A=C[/math] and[math]B=0[/math], the equation represents a circle, which is a special case of an ellipse;if [math]B2−4AC=0[/math], the equation represents a parabola;if [math]B2−4AC>0[/math], the equation represents a hyperbola;if we also have [math]A+C=0[/math], the equation represents a rectangular hyperbola.In the notation used here, [math]A[/math]and [math]B[/math] are polynomial coefficients, in contrast to some sources that denote the semi major and semi minor axes as [math]A[/math] and [math]B[/math].So, main difference between this 4 curves are their eccentricity.if[math]e=0[/math], the equation represents a circleif [math]0
What does the equation x^2 + y^2 -2xy +3x + 2 = 0 represent? Is it a parabola, ellipse, or hyperbola?
Because I’m lazy, I just used Desmos(Desmos | Beautiful, Free Math ):Anyway, we can see that the function’s tails go off into [math]-\infty[/math](this is also easily verifiable through analytic means), so I would be inclined to call it a parabola.
The graph of y=(x+50)^2-279 is obtained by performing the following transformations to the graph of y=x^2?
(I'm not exactly sure what the question is, but I suspect it is asking to identify the slide transformation, and express it algebraically?) y = (x + 50)² – 279 is a slide transformation of the standard parabola y = x² The slide takes all points P(x, y) ➞ P'(x – 50, y – 279) The slide image of the vertex V(0, 0) is V' (– 50, – 279).