+816 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)?
+128474 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)
+128474 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
