SOLUTION: Find the rule, in general form of a function with the unique zero -2 and passing through the point P(-1.3)

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Question 1127363: Find the rule, in general form of a function with the unique zero -2 and passing through the point P(-1.3)
Found 3 solutions by josgarithmetic, MathLover1, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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Find the rule, in general form of a function with the unique zero -2 and passing through the point P(-1.3)
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y=a%28x-%28-2%29%29
-
y=a%28x%2B2%29
3=a%28-1%2B2%29
3=a%281%29
a=3
-
y=3%28x%2B2%29
y=3x%2B6
-
highlight%28f%28x%29=3x%2B6%29

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
Find the rule, in general form of a function with:
the unique zerox%5B1%5D=+-2 and
passing through the point P(-1,3)
y=a%28x-x%5B1%5D%29
y=a%28x-%28-2%29%29
y=a%28x%2B2%29
use the point P(-1,3)
3=a%28-1%2B2%29
3=a%281%29
3=a
equation is:
y=3%28x%2B2%29
y=3x%2B6




or, you have this one too:
y=a%2Asqrt%28%28x-x%5B1%5D%29%29
y=a%2Asqrt%28x-%28-2%29%29
y=a%2Asqrt%28x%2B2%29
use the point P(-1,3%7D%7D%29+%0D%0A%0D%0A%7B%7B%7B3=a%2Asqrt%28-1%2B2%29
3=a%2Asqrt%281%29
3=a%281%29
3=a
y=3%2Asqrt%28x%2B2%29






Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Although you have received two responses to your question from other tutors, in fact the question you asked can't be answered. There are an infinite number of functions that have a unique zero of -2 and have graphs passing through (-1,3).

Both tutors showed you the unique linear function that satisfies both conditions, y+=+3x%2B6; that is the simplest such function.

One of the tutors also showed you the function y+=+3%2Asqrt%28x%2B2%29, which also satisfies the given conditions.

But the topic you chose for your question was quadratic functions. Perhaps you wanted a quadratic function answer....

In fact, there is also a unique quadratic function that satisfies both conditions. The root at -2 has to be a double root, which means the vertex of the graph is at (-2,0). The quadratic function with vertex (-2,0) passing through (-1,3) is

y+=+3%28x%2B2%29%5E2.

But there are infinitely many other functions that satisfy the given conditions; so in fact there is no "general form" of such a function.

Here is one polynomial function that satisfies the given conditions:

y+=+%283%2F2%29%28x%2B2%29%28x%5E2%2B1%29

The quadratic factor has no real roots, so the unique (real) root is at -2.

And by replacing the quadratic factor shown with a different one which has no real solutions, and probably changing the constant factor, you can get other polynomial functions that satisfy the given conditions.

So, without a more precisely worded question from you, the question really can't be answered.

Just for curiosity, here is a graph showing the three functions you have been shown, all of which have a unique root at -2 and whose graphs pass through the point (-1,3).