Question

Quadratic Functions. Solve each by graphing.
A) y=x^2+2x-3
B) y=-2x^2-8x+10

Answer 1

You can use the Desmos Graphing online feature to draw these graphs. The solutions are the values of x where the graphs cut the x-axis.

Answer 2

I have to show how i got it though.....How i got which points were on the graph @welshfella

Answer 3

question A)
hint:
I rewrite the quadratic function as follows:
\[\begin{gathered}
y = {x^2} + 2x - 3 \hfill \\
y = {x^2} + 2x + 4 - 4 - 3 \hfill \\
y = {\left( {x + 2} \right)^2} - 7 \hfill \\
y + 7 = {\left( {x + 2} \right)^2} \hfill \\
\end{gathered} \]

Answer 4

Now, if I make this traslation:
\(Y=y+7\), \(X=x+2\), where \(X,Y\) are the new coordinates, then I get:
\(Y=X^2\)

Answer 5

oh ok!!

Answer 6

The graph of the function of \(Y=X^2\), is:

Now, we have to understand where is located the origin of the coordinate system \(X,Y\)

Answer 7

In order to do that, we set \(X=0,Y=0\) into my traslation above:
\[\left\{ \begin{gathered}
Y = y + 7 \hfill \\
X = x + 2 \hfill \\
\end{gathered} \right.\]
So I get:
\[\left\{ \begin{gathered}
0 = y + 7 \hfill \\
0 = x + 2 \hfill \\
\end{gathered} \right.\]
please solve for \(x,y\)

Answer 8

oops.. sorry I have made a typo, here are the right formulas:
\[\begin{gathered}
y = {x^2} + 2x - 3 \hfill \\
y = {x^2} + 2x + 1 - 1 - 3 \hfill \\
y = {\left( {x + 2} \right)^2} - 4 \hfill \\
y + 4 = {\left( {x + 1} \right)^2} \hfill \\
\left\{ \begin{gathered}
Y = y + 4 \hfill \\
X = x + 1 \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \]
so please solve this system with respect to \(x,y\):
\[\left\{ \begin{gathered}
0 = y + 4 \hfill \\
0 = x + 1 \hfill \\
\end{gathered} \right.\]

Answer 9

please read the third step as follows:
\[y = {\left( {x + 1} \right)^2} - 4\]

Answer 10

ok