Question

Find the points where the parabola y=x+4x+3 intersects with the straight line y=-x+9

Answer 1

To find the points of intersection between the parabola y=x^2+4x+3 and the line y=-x+9, we need to set them equal to each other and solve for x.

Setting the equations equal, we have:

x^2 + 4x + 3 = -x + 9

Combining like terms:

x^2 + 5x + 6 = 0

Now, we can factor the quadratic equation:

(x + 2)(x + 3) = 0

Setting each factor equal to zero:

x + 2 = 0 or x + 3 = 0

Solving for x in each equation:

x = -2 or x = -3

Therefore, the parabola y=x^2+4x+3 intersects the line y=-x+9 at x=-2 and x=-3.

To find the corresponding y-values, substitute these x-values back into either equation:

For x = -2:
y = -(-2) + 9 = 2 + 9 = 11

For x = -3:
y = -(-3) + 9 = 3 + 9 = 12

Therefore, the points of intersection are (-2, 11) and (-3, 12).

Answer 2

x^2 + 5x - 6 = 0 because 3 - 9 = -6
(x-1)(x+6) = 0
x = 1 or x = -6
etc

Answer 3

Apologies for the mistake in the previous response. You are correct.

Let's solve the equation correctly this time.

Setting the equation x^2 + 5x - 6 = 0.

We can factor the quadratic equation as:

(x - 1)(x + 6) = 0.

Setting each factor equal to zero:

x - 1 = 0 or x + 6 = 0.

Solving for x in each equation:

x = 1 or x = -6.

Therefore, the parabola y = x^2 + 4x + 3 intersects the line y = -x + 9 at x = 1 and x = -6.

To find the corresponding y-values, substitute these x-values back into either equation:

For x = 1:
y = -1 + 9 = 8.

For x = -6:
y = 6 + 9 = 15.

Therefore, the points of intersection are (1, 8) and (-6, 15).