For what values of a and b is the line 2x + y = b tangent to the parabola y = ax2 when x = 5 a b

A quadratic equation is a polynomial equation of degree 2 .  The standard form of a quadratic equation is

0 = a x 2 + b x + c

where   a , b and c are all real numbers and a ≠ 0 .

If we replace 0 with y , then we get a quadratic function

      y = a x 2 + b x + c

whose graph will be a parabola .

The axis of symmetry of this parabola will be the line x = − b 2 a . The axis of symmetry passes through the vertex, and therefore the x -coordinate of the vertex is − b 2 a . Substitute x = − b 2 a in the equation to find the y -coordinate of the vertex. Substitute few more x -values in the equation to get the corresponding y -values and plot the points. Join them and extend the parabola.

Example 1:

Graph the parabola y = x 2 − 7 x + 2 .

Compare the equation with y = a x 2 + b x + c to find the values of a , b , and c .

Here, a = 1 , b = − 7 and c = 2 .

Use the values of the coefficients to write the equation of axis of symmetry .

The graph of a quadratic equation in the form   y = a x 2 + b x + c has as its axis of symmetry the line x = − b 2 a . So, the equation of the axis of symmetry of the given parabola is x = − ( − 7 ) 2 ( 1 ) or x = 7 2 .

Substitute x = 7 2 in the equation to find the y -coordinate of the vertex.

y = ( 7 2 ) 2 − 7 ( 7 2 ) + 2         = 49 4 − 49 2 + 2         = 49   −   98   +   8 4           = − 41 4

Therefore, the coordinates of the vertex are ( 7 2 , − 41 4 ) .

Now, substitute a few more x -values in the equation to get the corresponding y -values.

Plot the points and join them to get the parabola.

Example 2:

Graph the parabola y = − 2 x 2 + 5 x − 1 .

Compare the equation with y = a x 2 + b x + c to find the values of a , b , and c .

Here, a = − 2 , b = 5 and c = − 1 .

Use the values of the coefficients to write the equation of axis of symmetry.

The graph of a quadratic equation in the form   y = a x 2 + b x + c has as its axis of symmetry the line x = − b 2 a . So, the equation of the axis of symmetry of the given parabola is x = − ( 5 ) 2 ( − 2 ) or x = 5 4 .

Substitute x = 5 4 in the equation to find the y -coordinate of the vertex.

y = − 2 ( 5 4 ) 2 + 5 ( 5 4 ) − 1         = − 50 16 + 25 4 − 1         = − 50   +   100   −   16 16           = 34 16           = 17 8

Therefore, the coordinates of the vertex are ( 5 4 , 17 8 ) .

Now, substitute a few more x -values in the equation to get the corresponding y -values.

Plot the points and join them to get the parabola.

Example 3:

Graph the parabola x = y 2 + 4 y + 2 .

Here, x is a function of y . The parabola opens "sideways" and the axis of symmetry of the parabola is horizontal. The standard form of equation of a horizontal parabola is x = a y 2 + b y + c where a , b , and c are all real numbers and   a ≠ 0 and the equation of the axis of symmetry is y = − b 2 a .

Compare the equation with x = a y 2 + b y + c to find the values of a , b , and c .

Here, a = 1 , b = 4 and c = 2 .

Use the values of the coefficients to write the equation of axis of symmetry.

The graph of a quadratic equation in the form   x = a y 2 + b y + c has as its axis of symmetry the line y = − b 2 a . So, the equation of the axis of symmetry of the given parabola is y = − 4 2 ( 1 ) or y = − 2 .

Substitute y = − 2 in the equation to find the x -coordinate of the vertex.

x = ( − 2 ) 2 + 4 ( − 2 ) + 2         = 4 − 8 + 2         = − 2

Therefore, the coordinates of the vertex are ( − 2 , − 2 ) .

Now, substitute a few more y -values in the equation to get the corresponding x -values.

Plot the points and join them to get the parabola.

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To solve for the two variables #a# and #b#, we require a system of two equations. (Remember those? Well, they're coming in handy here!)

In order for two curves to be tangent to each other at a given point, they must have the same tangent line at that point. (If their tangent lines at the given point were different, the two curves would "pass through" each other rather than just "graze" each other.)

The slope of the tangent line to #y=x^3# at the point #(x_0, y_0)=(1,1)# is the first derivative of #y# (that is, #y'#) at #x_0=1#:

#y'=3x_0^2=3(1)^2=3#

Thus, the slope-point form of the tangent line required is:

#y-y_0=m(x-x_0)#
#y-1=3(x-1)#

For #y=x^2+ax+b# to share a tangent line at the point #(1,1)#, all we need is for the two slopes to be identical. (That way, the two tangent lines will have the same slope-point form).

Thus, we need #y=x^2+ax+b# to satisfy two conditions:

1. It must pass through #(x_0,y_0)=(1,1).#
2. It must have a slope (#y'#) of #3# at this point.

These will give us our two equations in two unknowns. The first one gives us

#color(white)=>y_0=x_0^2+ax_0+b#
#=>1=(1)^2+a(1)+b#
#=>0=a+b#

and the second one gives us

#color(white)=>y'=2x_0+a#
#=>3"  "=2(1)+a#
#=>a=1#

Looks like we know exactly what #a# is. All we need is to plug this value into the first equation and we'll find #b#:

#a+b=0" "<=>" "b="–"a#
#"                      "=>"    "b="–"1#

Here is a graph of the two curves:
graph{(y-x^3)(y-x^2-x+1)=0 [-4.937, 4.93, -1.716, 3.22]}

Bonus:

The quick way to solve this problem is to realize that both derivatives must be equal at #(1,1)#, so we equate the two as follows:

#color(white)=>3x^2=y'=2x+a#

#=>3(1)^2=2(1)+a#

#=>3-2=a" "=>" "a=1#

Then, we just have to plug in the known values for #x, y#, and #a# into #y=x^2+ax+b# to find #b#:

#color(white)=>y=x^2+ax+b#

#=>1=(1)^2+(1)(1)+b#

#=>1-2=b" "=>" "b="–"1#.

2 Answers By Expert Tutors