# How do you solve #y=-x^2+2x-3# and y=x-5?

(-1, - 6)

(2, - 3)

This is a downward facing U shaped curve

y = x - 5 -------------------------(2)

Straight line cuts it at two points.

Substitute

x ( - x + 2) + 1 (- x +2) = 0

(x + 1) ( - x + 2) = 0

(x + 1 = 0

x = -1

- x + 2 = 0

x = 2

Substitute the x values in equation (1)

At x = - 1 ; y = -1 - 5 = - 6(- 1, -6) one point

At x = 2 ; y = 2 - 5 = 3

(2, -3) is another point

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To solve the system of equations (y = -x^2 + 2x - 3) and (y = x - 5), set the equations equal to each other:

[ -x^2 + 2x - 3 = x - 5 ]

Rearrange the equation to set it equal to zero:

[ -x^2 + 2x - 3 - x + 5 = 0 ]

[ -x^2 + x - 2 = 0 ]

Now, solve the quadratic equation using factoring, completing the square, or the quadratic formula. Let's use factoring:

[ -(x^2 - x + 2) = 0 ]

[ -(x - 2)(x + 1) = 0 ]

Setting each factor equal to zero:

[ x - 2 = 0 ] or [ x + 1 = 0 ]

Solving for (x):

[ x = 2 ] or [ x = -1 ]

Now, substitute the values of (x) into one of the original equations to find the corresponding (y) values. Let's use (y = -x^2 + 2x - 3):

When (x = 2):

[ y = -(2)^2 + 2(2) - 3 = -4 + 4 - 3 = -3 ]

When (x = -1):

[ y = -(-1)^2 + 2(-1) - 3 = -1 - 2 - 3 = -6 ]

So, the solutions to the system of equations are the points ((2, -3)) and ((-1, -6)).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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