Question

For what value of h will the line y=-2x+h be tangent to y=3x²+4x-1?

b) Graph the line and the curve on the same set of axes to verify the solution. Use a scale.

 

Answer 1

$\text{Given: }y=-2x+h\text{ ; }y=3x^2+4x-1$

 

$\text{Step 1: Equate the two equations. }$

$-2x+h=3x^2+4x-13x^2+4x+2x-h-1=03{\mathbf{x}}^2+6\mathbf{x}-(\mathbf{h}+1)=0$

 

$\text{Step 2: Use discriminant equation and equate it to zero. }$

$3x^2+6x-(h+1)=0a=3\text{ ; }b=6\text{ ; }c=-(h+1)D=b^2-4ac{b}^2-4ac=0$                

 

$\text{Step 3: Solving for h.}$

$b^2-4ac=0(6{)}^2-4(3)[-(h+1)]=036+12h+12=012h=-48\mathbf{h}=-4$

 

So the equation of the line tangent to the parabola of y=3x2+4x-1 is  y=-2x-4.

 

Discriminant, D is the determination is how many possible solution a quadratic equation has. When D=0 the quadratic equation will have 1 solution. So by equating the line and the parabola a quadratic equation is formed and the value of h is solve and this will give the point of tangency.

So, the equation of the line tangent to the parabola of y=3x2+4x-1 is  y=-2x-4.

Therefore, h=4