Question

Question #c4080

Answer 1

`y=-25/8`

Explanation:

`y = 2x^2-x-3`

Firstly we note that at a vertex (max/min) the tangent will be horizontal (`y`=constant), so we just need to find the coordinates of the max/min. There are two ways to do this either with or without calculus:

Method 1 (pre-calculus):
Complete the square:
`y = 2x^2-x-3`
`:. y = 2{x^2-x/2-3/2}`
`:. y = 2{(x-(1/4))^2 -(1/4)^2-3/2}`
`:. y = 2{(x-1/4)^2 -1/16-3/2}`
`:. y = 2{(x-1/4)^2 -25/16}`
So a minimum occurs when `x=1/4`

Method 2 (Calculus):
`y = 2x^2-x-3`
differentiating wrt `x` gives;
`dy/dx = 4x-1`
At a max/min `dy/dx=0 => 4x-1=0`
`:. x=1/4` (as above).
We observe that as the function is a positive quadratic it will have a U shared curve, so this corresponds to a minimum

Finding the tangent:
So back to business, we now know that the min occurs when `x=1/4`
`x=1/4=>y=2(1/16)-1/4-3`
`:. y=1/8-1/4-3`
`:. y=-25/8`

So the equation of the tangent at the vertex (corresponding to `x=1/4`) is `y=-25/8`