### Instruction: Two parabolas are the graphs of the equations $y=2x^2-10x-10$ and $y=x^2-4x+6$. Find all points where they intersect. List the points in order of increasing $x$-coordinate, separated by semicolons.### Response: To find the points of intersection, I need to solve the system of equations $y=2x^2-10x-10$ and $y=x^2-4x+6$.
That means I need to find the values of $x$ and $y$ that make both equations true at the same time.
One way to do that is to set the two expressions for $y$ equal to each other and solve for $x$.
That will give me a quadratic equation in $x$.
So I have $2x^2-10x-10=x^2-4x+6$.
To simplify this equation, I can subtract $x^2-4x+6$ from both sides.
That gives me $x^2-6x-16=0$.
Now I need to factor this equation or use the quadratic formula to find the roots.
I notice that this equation can be factored as $(x-8)(x+2)=0$.
So the roots are $x=8$ and $x=-2$.
These are the $x$-coordinates of the points of intersection.
To find the corresponding $y$-coordinates, I can plug these values of $x$ into either of the original equations for $y$.
I'll use the first one, $y=2x^2-10x-10$.
When $x=8$, I have $y=2(8)^2-10(8)-10=2(64)-80-10=128-90=38$.
So one point of intersection is $(8,38)$.
When $x=-2$, I have $y=2(-2)^2-10(-2)-10=2(4)+20-10=8+10=18$.
So another point of intersection is $(-2,18)$.
These are the only two points of intersection, since a quadratic equation can have at most two real roots.
To list them in order of increasing $x$-coordinate, I start with the one that has the smaller $x$-value, which is $(-2,18)$.
Then I write a semicolon and write the other point, which is $(8,38)$.