Question

Y=2x^2-3x+3, y=3-x^2

Answer 1

2х^2-3х+3=3-х^2

2х^2-3х+3-3+х^2=0

3х^2-3х=0

3х(х-1)=0

1). 3х=0; х=0

2). х-1=0; х=1

Answer 2

To find the points of intersection between the two curves represented by the equations $Y=2x^2-3x+3$ and $y=3-x^2$, we need to set them equal to each other and solve for the value(s) of $x$ that make them true.

So, we have:

$2x^2 - 3x + 3 = 3 - x^2$

Let's rearrange the equation and set it to zero to solve for $x$:

$2x^2 - 3x + 3 - (3 - x^2) = 0$

$2x^2 - 3x + 3 - 3 + x^2 = 0$

Combine like terms:

$3x^2 - 3x = 0$

Factor out $3x$ from the left side:

$3x(x - 1) = 0$

Now, we have two possible solutions for $x$:

1. $3x = 0 \Rightarrow x = 0$
2. $x - 1 = 0 \Rightarrow x = 1$

Now, we can find the corresponding values of $y$ for each value of $x$:

For $x = 0$:

$y = 3 - (0)^2 = 3$

For $x = 1$:

$y = 3 - (1)^2 = 3 - 1 = 2$

So, the two curves intersect at two points: (0, 3) and (1, 2).