show that y=3x^2+6x+3 and y=4x+3 intersect. state the point of intersection. how do i do this?

Question

please help with this question above.

Answer 1

Find the points of intersection by algebraically.

The quadratic equation y = 3x ² + 6x + 3

The linear function y  = 4x + 3

the linear function is already solved for one y .

Substituting the expression for y  into the quadratic function, we get

3x ² + 6x + 3 = 4x + 3

3x ² + 6x + 3 - 4x -3 = 0

3x ² + 2x  = 0

Now we solve the resulting quadratic equation for x .

x(3x + 2) = 0

x = 0 and 3x + 2 = 0

x = 0 and x = -2/3

There are two solutions to the quadratic equation: x = 0, and x  = -2/3.

Substituting each of these solutions into either of the two original functions (the linear one would be easier) leads us to the corresponding y  - values.

y  = 4x + 3

For x = 0

y  = 4(0) + 3

y  = 3

For x = -2/3

y  = 4(-2/3) + 3

y  = -8/3 + 3

y  = (-8+9)/3

y  = 1/3

Thus, the two points of intersection are (0 , 3) and (-2/3 , 1/3).

Find the points of intersection by graphically.

y = 4x + 3

Compare it to y = mx + b  is slope intercept form of a line.

y = 3x ²+ 6x + 3.

Compare it to standard form of parabola y = ax ²+ bx + c .

1. Draw coordinate plane.

2. Graph thethe line y = 4x +3 by using slope - intercept form.

3.Draw the parabola y = 3x ² + 6x + 3.

Points of intersection are (0 , 3) and (-2/3 , 1/3).