Draw a possible function that satisfy the given domain and range
1 answer:
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Answer:
see attached.
Step-by-step explanation:
Your graph demonstrates a possible function, but it has the domain and range swapped.
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The <em>domain</em> is the <em>horizontal</em> extent, so your line must end at x=-9 on the left and x=6 on the right.
The <em>range</em> is the <em>vertical</em> extent, so the function must extend from -4 at the bottom to +8 at the top.
The attachment shows a linear function with those characteristics.
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The parabola will show the vertex in the format: y-k = (x-h)^2, where the vertex point
lies at (h, k).
let's first put it in "y =" standard format:
Since we cannot get a perfect square out of this, we complete the square: a=1, b=2, c=3
(b/2)^2 = (2/2)^2 = 1, so
So there's +2 leftover, since 3-1=2; so:
Now we'll subtract the 2 from both sides to show our vertex:
where our vertex (h, k) is at (-1, 2)
lies at (h, k).
let's first put it in "y =" standard format:
Since we cannot get a perfect square out of this, we complete the square: a=1, b=2, c=3
(b/2)^2 = (2/2)^2 = 1, so
So there's +2 leftover, since 3-1=2; so:
Now we'll subtract the 2 from both sides to show our vertex:
where our vertex (h, k) is at (-1, 2)