A quadratic function is defined by y= 3x^2 + 4x - 2. A linear function is defined by y = mx-5. What values of the slope of the line would make it tangent

Question

im in high school and I haven’t learned derivetives yet so it would be great if the answer was able to be solved as simple as possible

Answer 1

First, we find the vertex.  If the vertex has the same y-coordinate as the given line, then the slope of the line would be zero.

 

y = 3(x² + (4/3)x + 4/9) - (4/3) - 2

 

y = 3(x + 2/3)² - 10/3

 

The y coordinate of the vertex is not -5.  Therefore, we verified that the slope of the line is not zero.

 

 

So now, we find the derivative of the parabola using the power rule or limit definition.

 

y' = 6x + 4

 

This is the slope of the tangent line.

 

Next, we find the points of tangency by setting the line and parabola equal to each other.

 

3x² + 4x - 2 = (6x + 4)x - 5

 

3x² + 4x - 2 = 6x² + 4x - 5

 

 

Bring all terms to the right side of the equation.

 

0 = 3x² - 3

 

Factor the quadratic equation.

 

0 = 3(x² - 1)

 

Then        

 

x = -1      and        x = 1

 

 

Next, we plug in the values of x back into the slope.  The slope was the derivative we found in terms of x.

 

slope = 6(-1) + 4 = -2

slope = 6(1) + 4 = 10

 

These are your values of the slope.