For the equation y=a(x-h)squared+k...?

Question

Explain how you know which way the parabola opens.
Explain how to find the y-intercept
Explain how to find the x and y value of the vertex.

Don't have to answer them all! Answering any is appreciated!

Answer 1

for the first part...if a is a negative value then the parabola will be drawn downwards (somewhat like a sad face) and likewise if the a value is positive the parabola will be drawn upwards like a smiley face.

to find the y intercept substitute x=0 into the equation...so the y intercept is equal to a multiplied by the square of negative h plus k.

the turning point is represented by (h,k) the x value of the turning point is equal to h and the y value is equal to k.

hope that helped.

Answer 2

Given the equation y = a(x - h)^2 + k. Then, the graph of this equation in the Cartesian plane is a parabola wherein it is either opening upward or downward with respect to its vertex whose coordinate is (h, k).

The way this parabola opens depends upon the non-zero value of a. That is:
(a) if a > 0, the graph of the parabola opens upward; and
(b) if a < 0, the graph of the parabola opens downward.
If we have a = 0, the given equation reduces to y = k, in which its graph is a horizontal line; which is a degenerate case of a parabola.

Next, to obtain the y-intercept, we simply set the x = 0. That is:
y = a(0 - h)^2 + k = ah^2 + k.

This means that given the equation y = a(x - h)^2 + k, its graph will intersect at y-axis exactly at the point (0, ah^2 + k).

Finally, observe that the given equation can be written in standard form as follows:
(y - k) = a(x - h)^2,
in which the coordinate (h, k) is the vertex of the given parabola.

Answer 3

For large values of x, y gets large (rises) and for small values of x (negative values) y also gets large (rises) so the parabola opens upward. The y-intercept is the value of y when x = 0. so in this case that would be a*h^2 + k. The vertex is where the slope of the curve is zero; it is also the minimum value of y. Both are found the same way: take the derivative of y and set it to zero. The derivative is 2*a*(x-h); it is zero at x = h; the value of y at x=h is k.

Answer 4

The form of the equation tells you what kind of parabola (or graph) it is, but only after some practice with differents kinds you learned in class. Just remember, the formula is just a number cruncher with a pattern. To find the y int, or (x,y) or the vertex, you'd need some numbers to plug in. To find the y, you have to solve for y. So y would be on the left side of the equation, with all your work on the right side. The (x,y)... man that's tougher. I think either you have to break it into binomials, or you know the graph well enough to see it in your head. I think I did both, but it was years ago. Either way, again, you need #'s so plug in.

Answer 5

to know the shape of the parabola u-shaped or n-shaped,look at the value of 'a' in the equation you gave.if 'a' is greater than zero(positive value) the shape of the graph is u-shaped with minimum point.if 'a'is les than zero(negative value)it has n-shaped graph with maximum point.


i cannot write more because of the limited funtion.but you should visit this website.

www.okc.cc.ok.us/maustin/Quadratic_Functions/Quadratic%20Functions.htm

Answer 6

since the equation is of the form x^2=y it will open eithe upwards or downeards
since a=positive it will open upwards
for y intercept put x=0
the vertex=h,k with th signs in the given form of the equation