find, in the form y=ax^2+bx+c, the equation of the quadratic whose graph touches the x axis at 4 and passes through (2,12)

Question

Steps PLEASE

Answer 1

Plug the points you have (2,12) and (4,0) (from the x intercept) into 2 separate equations

12=a (2)²+b (2)+c

12=4a+2b+c

 

0=a (4)²+b (4)+c

0=16a+4b+c

 

Now you want to get rid of one variable to start to solve using these 2 equations, and c is going to be the easiest. So we will multiply the first equation by -1, and add the 2 equations together.

 

-12=-4a-2b-c

+ 0=16a+4b+c

 

The c's cancel out and we are left with

-12=12a+2b

Divide by 2 to get b by itself

-6=6a+b

Get other variables on the other side

B=-6a-6

 

Plug back in to one of your equations to find C

12=4a+2 (-6a-6)+c

12=4a-12a-12+c

12=-8a-12+c

24=-8a+c

C=8a+24

 

Plug one last time with all A variables

12=4a-12a-12+8a+24

Simplify

12=12 

Because 4-12+8=0 and -12+24=12

This means a=1

 

Knowing this, go back to where we found the individual values of b and c.

B=-6a-6

C=8a+24

Plug in 1 for your A value

B=-6 (1)-6

B=-12

 

C=8 (1)+24

C=32

 

So your final equation is

Y=x²-12x+32

 

It intercepts the x axis at (4,0) and passes through (2,12)

 

 

Answer 2

We have three pieces of data 

1. (2,12) is a point on the curve. 

plugging that into the equation we have

4a + 2b + c = 12 (I)

 

2. (4,0) is a point on the curve

plugging that into the equation we have  

16a + 4b + c = 0 (II)

 

3. y = 0 is a tangent at (4,0)  (since it "touches" x axis)

The slope of the tangent is given by m= y' = 2ax +b

Since we are interested in tangent at x=4 the slope will be 8a+b. This must equal 0 since the tangent itself is y=0.

8a + b = 0 (III)

=> b = -8a

 

plugging this into (II)

we have

16a -32a +c = 0

=> c = 16a

 

Plugging values for b and c in (I)

4a -16a + 16a = 12

=> 4a = 12

=> a = 3

=> b = -24

=> c = 48

 

Therefore y = 3x^2-24x +48