We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
206
3
avatar+814 

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+101796 
+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+101796 
+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+814 
+1

Great solutions, CPhill! Thank you!

 Aug 12, 2018

10 Online Users

avatar