How to Find Domain and Range of Parabola with no graph?

Question

I have been reviewing for my grade 10 math exam, but got confused when I was given 5/4(x+2)^2 -1 and was asked for the domain and range of it. I know how to find the vertex, x-intercepts and y-intercept, but how do you find domain and range if you aren't given a graph of the parabola? Thanks

Answer 1

every parabola has a domian of x = all real numbers
the range of a parabola is determine by the y value of hte vertex and whether the porabola opens up or down.

Vertex form:
f(x) = a(a - h)^2 + k with the vertex of (h,k)

f(x) = 5/4 (x + 2)^2 - 1 has the vertex of (-2,-1)

because 5/4 is a positive number, the parabola opents up, U shape, and when the parabola opens up, it has a minumber point, which is the vertex of the parabola. The minimun point is (-2,-1). Because we are looking for the range, we want to look at t y value or the minimun point, which is -1. The graphs can not be lower a than -1, thus, the range is y >/ -1.


To find y-intercept, plug 0 for x, and solve
To find x-intercept, plug 0 for y, and solve


hope this helps

Answer 2

5/4(x+2)^2 -1
It would be helpful if you made you equations unambiguousIs the 5 divided by just the 4 ,or is it divided by 4(x+2)^2? It makes a big difference which it us.
If the equation is 1.25*(x+2)^2 -1, then the domain is all real values of x. This is because a parabola with its axis of symmetry parallel to the y-axis is a continuous function.

This parabola has a minimum value because the coefficient of x is positive. You have found the vertex so you know the value of y which is the value of y at the vertex. So the range is all real values >y vertex.

Answer 3

A parabola in the form of (y-k) = 4p(x - h)^2 (meaning you have the x term squared and not the y term) is going to have the vertex at (h,k) and will have one of two different shapes: opening upwards, or opening downwards. If p is negative, then it opens downwards. If p is positive (as it is in your example) then it opens upwards.

So the range depends on where the vertex is and where it curves. If the parabola opens downwards, then the range is y <= k. If it opens upwards, then the range is y >=k.

The domain is going to be the set of all real numbers, because x is the independent variable. This is regardless of whether the parabola points up or down.

Answer 4

The domain is simply all values of x over whick the function is deffined. In this case you can plug in any value of x and get a unique value of y so the domain covers all real numbers.
The range is all values of y for which the function is deffined. You will be able to find this using the vertex. We see that the lowest value y can ever be is -1 when x= -2. So the range is all real numbers greater than or equal to -1

Answer 5

The domain is all real #'s. There are no values of x that cause a problem.

The range is easy once you find the vertex. If v is the y value of the vertex (-1 in your example), then the range is y>=v or y<=v, depending on the sign (+ or -) of the coefficient of the quadratic term. In your example the range is y >= -1

Answer 6

well, for the first part, you said that y/x^2 = 1; here, you would be solving for when y divided by x squared is equal to one, so you would use a pair, and the answer would not include a 0 for the x-coordinate. The same thing goes for the second part. Basically, you wouldn't graph x/y^2=(a number); that would be solving for a point

Answer 7

How To Find The Range