Find the equation of the line through the origin and the minimum point of the curve y=3x^2-12x?

Question

What do they exactly mean in this question

Answer 1

Alright it's getting pretty late but nobody has answered your question yet.

It is asking for the equation of the line (y=mx+b) that travels through (0,0) and the minimum point on the curve y= 3x^2-12x

Since I am low on time I will solve it and write a guide for you to solve it if you have the right materials, which is a Texas Instruments graphing calculator.


Press "Y=" and type in "3x^2-12x"
Hit "Graph" and the parabola should appear
Press "2nd" -"Calc"- "minimum"
Go to the left bound then right and hit "enter"

Your minimum point on the curve is (2,-12)

Hit "STAT" - "Edit"
Clear the table if necasary
Under L1 type "0" "enter" "2"
Under L2 type "0" "enter" "-12"
Hit "Stat" - "Calc" - "LinReg(ax+b)"
Hit enter twice and it will give you:

y=ax+b
a= -6
b=0


This means your final answer is y = -6x
If you are quick and experienced it might take you all of 20 seconds to complete.

Hope this helps.

Answer 2

Find the equation of the line through the origin and the minimum point of the curve y=3x^2-12x.
What do they exactly mean in this question?

You need to find the equation two points. One is the origin and the other is the vertex of the parabola.

Find the vertex of the parabola by completing the square.

y = 3x² - 12x = 3(x² - 4x)
y + 3*4 = 3(x² - 4x + 4)
y + 12 = 3(x - 2)²

The vertex (h,k) = (2, -12)

Calculate the slope of the line thru (0,0) and (2,-12).

m = ∆y/∆x = (0 - -12) / (0 - 2) = 12 / -2 = -6

Plug in one of the points and write the equation of the line.

y - 0 = -6(x - 0)
y = -6x