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- anonymous

What is the vertex form of the equation?
y=x^2+4x-3
A y=(x-2)^2-7
B y=(x+2)^2-7
C y=(x-2)^2+7
D y=(x+2)^2+7
Please explain how to do this

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- anonymous

- schrodinger

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- triciaal

|dw:1444080552670:dw|

- whpalmer4

We have \[y = x^2+4x-3\]and we need to get it in vertex form. Whenever we have an equation of the form \[y = ax^2+bx+c\]where \(a,b,c\) are constants and \(a\ne0\), we have a parabola.
\[y = x^2+4x-3\]
Notice that this can be rewritten\[y+3 = x^2+4x-3+3\]\[y+3 = x^2+4x\]
At this point, notice that what we have on the right hand side is very similar to \[(x+2)^2 = (x+2)(x+2) = x^2+2x+2x+2*2=x^2+4x+4\]Only difference is that we don't have that \(4\) at the end. But if we subtract \(4\) from both sides, we get this:
\[(x+2)^2 -4 = x^2+4x +4 - 4= x^2+4x\]which is what we have on our right hand side. Let's rewrite again using that:
\[y+3 = (x+2)^2 - 4\]Now let's shuffle that \(3\) over to the right hand side:\[y+3-3 = (x+2)^2-4-3\]\[y=(x+2)^2-7\]
Two things to note here:
1) we've got something that appears in the answer choices
2) this looks very much like a shifted function
If we have a function \(y = f(x)\), we can shift it up or down by adding a constant to it.
\[y = f(x) + 3\]gives us the same graph as \(y = f(x)\), except every point is shifted by 3 units along the positive \(y\)-axis. Similarly, \(y = f(x) - 2\) gives us the same graph as \(y = f(x)\) except this time each point is shifted by 2 units along the negative \(y\)-axis.
Now if we had a simple parabola, \(y = x^2\), the vertex is at \((0,0)\). If we write it as \(y = f(x) = x^2\), then we could shift that parabola up and down by adding a constant, let's call it \(k\):
\[y = f(x)+k = x^2 + k\]
Hmm, that looks a bit like our equation, doesn't it? Except we have \((x+2)^2\) instead of \(x^2\). Well, I said we can shift our parabola up and down by adding a constant; is it possible to shift it right and left by doing something else? Good question! Well, it turns out the answer is yes. Imagine if we drew our parabola on a piece of clear plastic, and just slid it to the right or left? That would be the same as adding or subtracting from the value of \(x\) BEFORE we plug it into the equation. \[y = f(x-1) = (x-1)^2\]makes a parabola that looks identical to \(y = f(x) = x^2\), but it is shifted 1 unit to the right. If we add to the value of \(x\) instead of subtracting, we shift it in the other direction along the \(x\)-axis.
This means our simple parabola could be written \[y = f(x-h) = (x-h)^2\] where \(h\) is how much we shifted it to the left.
So, if we look at tricia's equation, now we can see that she has given us the equation of a parabola, shifted left by \(h\) units and up by \(k\) units, and as the vertex of a parabola which has not been shifted at all is at \((0,0)\), the vertex of a shifted parabola will be at \((h,k)\)!
Let's try it out: how about a parabola with a vertex at \((2,1) = (h,k)\):
\[y = (x-h)^2+k = (x-2)^2+1 \]\[=x^2-2x-2x-2(-2)+1\]\[=x^2-4x+5\]
If we make a little plot from say \(x=-4\) to \(x=4\) we get the following points:
\[
\begin{array}{cc}
x&y\\\hline -1 & 10 \\
0 & 5 \\
1 & 2 \\
2 & 1 \\
3 & 2 \\
4 & 5 \\
5 & 10 \\
\end{array}
\]
If you plot those points on a piece of graph paper and run a smooth curve through them, you'll see that you have made a parabola with vertex at \((2,1)\), just like we set out to do.
I agree, this is more than you need to do the problem, but put here in the hope that it might foster some understanding of what is going on, rather than just giving you a formula which works "as if by magic!"

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