Parabola Help. (1 Viewer)

Lemiixem · Jul 16, 2012

The graph of a parabolic function crosses the x-axis at the origin and at x=4. If the minimum value of the function is -1, determine the parabolic function.

A right isosceles triangle has one vertex at the origin O and another at the point A(1,3). The base of the triangle has the equation x=2y.
Show that the third vertex B has coordinates (4,2)

Sy123 · Jul 16, 2012

Parabola form:

$y=ax^2+bx+c \\ y'=2ax+b \\ y=0 \ \ \Rightarrow \ \ x=0 \ \ x=4 \\ y'=0 \rightarrow \ \ y=-1 \ (x=2) \\ \\ \therefore 0=(4)^2a+4b+c \\ 0=0+0+c \\ \therefore c=0 \\ \therefore 16a+4b=0 \\ \\ pt(2, -1) \\ \\ -1=4a+2b \\ \\ \therefore a=\frac{1}{4} \ \ b=-1 \\ \\ \therfore \frac{x^2}{4}-x=y$

Easier way (but much more risky in my opinion)

$y=(x-a)(x-b) \\ $Roots at$ \ \ x=0 \ \ x=4 \\ \\ \therefore y=x(x-4)$

I think they want you to do the first way because they gave you the minimum value.

Will do second one later

Timske · Jul 16, 2012

should be 4y = x(x-4)

Lemiixem · Jul 16, 2012

Bump.
Could you guys please help me with this question.

3y=mx+6 and y=log(x+1) [Base e], the vertical distance from the straight line to the curve is given by mx/3 + 2 - log(x+1) [Base e]. Show that the shortest vertical distance is given the expression 3-m/3 - log(3/m) [Base e]

Sy123 · Jul 16, 2012

"the vertical distance from the straight line to the curve is given by"

Is that exactly what the question says? It doesnt really make sense.

Lemiixem · Jul 16, 2012

The question asked to draw the curve first, I'll type out the whole question.

B) Consider a straight line with the equation 3y=mx+6, M>0, and a curve with equation y=log(x+1)

(i) By Substituting m=2, sketch 3y=2x+6 and y=log(x+1) together on the same diagram.

(ii) Show that the vertical distance to the line is given by the expression 2x/3 + - log(x+1)

(iii) Now for the more general case where 3y=mx+6 and y=log(x+1), the vertical distance from the straight line to the cuvrve is given by mx/3 + 2 - log(x+1). Show that the shortest vertical distance is given the expression 3 - m/3 - log(3/m)

I managed to do the first two, just need help with (iii). All logs are base e.

Sy123 · Jul 16, 2012

Ok I get it, its asking the vertical line distance in terms of variable x.
Thats fine then, ok the vertical distance between variable points:
$(x,\frac{mx+6}{3}) \ \ \ (x, \ln(x+1))$

Is given by distance formula:

$D^2=(x-x)^2+(\frac{mx}{3}+2-\ln(x+1)) \\ \\ D=\frac{mx}{3}+2-\ln(x+1) \\ \frac{dD}{dx}=\frac{m}{3}-\frac{1}{x+1} \\ \\ 0=\frac{m}{3}-\frac{1}{x+1} \\ \\ x=\frac{3}{m}-1$

Minimum since there is no maximum distance. (We dont need to assume this I would think)

$\therefore $Minimum distance occurs at$ x=\frac{3}{m}-1 \\ \\ D=\frac{m(\frac{3}{m}-1)}{3}+2-\ln(\frac{3}{m}) \\ \\ D=3-\frac{m}{3}-\ln(\frac{3}{m})$

Skipped algebra, you get the concepts.

Parabola Help. (1 Viewer)

Lemiixem

Member

Sy123

This too shall pass

Timske

Sequential

Lemiixem

Member

Sy123

This too shall pass

Lemiixem

Member

Sy123

This too shall pass

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