SOLUTION: Hello I am graphing the vertex and axis of symmetry and I understand how to find it, but how about when the vertex is a fraction? How would I graph the vertex? (if x and y are in

Algebra ->  Graphs -> SOLUTION: Hello I am graphing the vertex and axis of symmetry and I understand how to find it, but how about when the vertex is a fraction? How would I graph the vertex? (if x and y are in       Log On


   



Question 1135161: Hello I am graphing the vertex and axis of symmetry and I understand how to find it, but how about when the vertex is a fraction? How would I graph the vertex? (if x and y are in fraction form)
Found 2 solutions by greenestamps, MathLover1:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


I don't see a problem... fractions graph the same way as whole numbers....

Perhaps you can re-post your question, being more specific about why you don't understand how to do what you are doing.

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You ask what if the coordinates of the vertex were (1/2,9/4).

The x coordinate, 1/2, is halfway between 0 and 1; the y coordinate, 9/4 = 2 1/4, is 1/4 of the way from 2 to 3. If you are graphing by hand, any good approximation to those positions should be acceptable.

Perhaps you are doing the graphing on a computer using a graphing tool. If so, and the coordinates of the vertex are fractions, then the tool certainly should let you have coordinates that are fractions; it wouldn't make sense to have them ask you to plot a vertex at (1/2,9/4) if the graphing tool only uses integer coordinates....

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
The axis of symmetry always passes through the vertex of the parabola . The -coordinate of the vertex is the equation of the axis of symmetry of the parabola. For a quadratic function in standard form, y+=+a+x%5E2+%2B+b+x+%2B+c , and in vertex form y=a%28x-h%29%5E2%2Bk the axis of symmetry is a vertical line x+=+-b%2F+2a
example:
y=x%5E2-x%2B1....use vertex form to find coordinates
y=%28x%5E2-x%2Bb%5E2%29-b%5E2%2B1
y=%28x%5E2-x%2B%281%2F2%29%5E2%29-%281%2F2%29%5E2%2B1
y=%28x-1%2F2%29%5E2-1%2F4%2B1
y=%28x-1%2F2%29%5E2-1%2F4%2B4%2F4
y=%28x-1%2F2%29%5E2%2B3%2F4
so, h=1%2F2 and k=3%2F4 and vertex is at (1%2F2, 3%2F4)