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1. If the parabola \(y_1 = x^2 + 2x + 7\) and the line \(y_2 = 6x + b\) intersect at only one point, what is the value of \(b\)?

 

2. The quadratic \(x^2+(2.6)x+3.6\) can be written in the form \((x+b)^2+c\), where \(b\) and \(c\) are constants. What is  \(b+c\) (as a decimal)?

 Aug 11, 2018
 #1
avatar+98129 
+2

1.  Set the equations equal

 

x^2 + 2x + 7  = 6x + b      simplify

x^2 - 4x + (7 - b)   = 0

 

If there is  a single solution to this, the discriminant must  = 0

Therefore

 

(-4)^2  - 4 (1) ( 7 - b)   = 0  simplify

16 - 4(7 - b)   = 0

16 - 28 + 4b  = 0

-12 + 4b  = 0

-12  = -4b    divide  both sides  by -4

3  = b

 

Here's a graph : https://www.desmos.com/calculator/fykdlia1ej....the intersection point is  (2, 15)

 

 

cool cool cool

 Aug 11, 2018
 #2
avatar+98129 
+2

2.  x^2 + (2.6)x + 3.6    .....complete the square  on  x

 

Take (1/2) of 2.6  = 1.3....square it  = 1.69 ....add it and subtract it.......

 

x^2 + (2.6)x  + 1.69 + 3.6 - 1.69        facto the  first three terms....simplify the rest

 

(x + 1.3)^2  + 1.91

 

b = 1.3   c  = 1.91

 

So

 

b + c  =  1.3  + 1.91   =  3.21

 

 

 

cool cool cool

 Aug 11, 2018
 #3
avatar+809 
+1

Great solutions, CPhill! Thank you!

 Aug 12, 2018

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