Question

For what value of b will the line y= -2x + b be tangent to the parabola y= 3x^2 + 4x - 1?

Answer 1

i hope this is a calculus course

Answer 2

the derivative of \(y=3x^2+4x-1\) is \(y'=6x+4\)

Answer 3

Not a calculus course :S

Answer 4

damn

Answer 5

ok then \(y\) has to satisfy two equations
\[y=-2x+b\] and \[y=3x^2+4x+1\] we can set
\[3x^2+x+1=-2x+b\] and solve

Answer 6

we get
\[3x^2+3x+(1-b)=0\]

Answer 7

in order for the line to be tangent, it must touch and not cross
that means the above equation can have only one solution
that means the discriminant must be zero

Answer 8

Hm, the answers says -4

Answer 9

all my algebra is wrong that is why
lets start again

Answer 10

\[3x^2+4x+1=-2x+b\]
\[3x^2+6x+(1-b)=0\]

Answer 11

now \[a=3,b=6,c=(1-b)\]

Answer 12

\[b^2-4ac=36-12(1-b)=0\]
\[36-12+12b=0\]
\[24+12b=0\]
\[24=-12b\]
\[b=-2\]damn still wrong

Answer 13

picking up from @satellite73 's line

\(\bf y = 3x^2+4x-1\qquad tangent\quad \implies y = -2x+b\\ \quad \\ \quad \\
3x^2+4x-1 \quad = \quad -2x+b\implies 3x^2+6x-1-b=0\\ \quad \\
3x^2+6x-(1+b)=0\implies 3x^2+6x+(-1-b)=0\\ \quad \\
\textit{as satellite73 said, let us equate the discriminant to 0}\\ \quad \\
b^2-4ac = 0\implies (6)^2-4(3)(-1-b)=0\implies 36-12(-1-b)=0\\ \quad \\
36+12+12b=0\implies 48+12b=0 \implies 12b=-48\implies b = -4\)