Question

What is the vertex of `y=-3x^2+5x+6`?

Answer 1

`0.833, 8.083`

Explanation:

The vertex can be found using differentiation, differentiating the equation and solving for 0 can determine where the x point of the vertex lies.

`dy/dx (-3x^2 + 5x +6) = -6x + 5`
`-6x + 5 = 0, 6x = 5, x = 5/6`

Thus the `x` coordinate of the vertex is `5/6`
Now we can substitute `x = 5/6` back into the original equation and solve for `y`.

`y = -3(5/6)^2 + 5(5/6) + 6`
`y = 8.0833`

Answer 2

`(5/6,97/12)`

Explanation:

`"for a parabola in standard form " y=ax^2+bx+c`

`"the x-coordinate of the vertex is " x_(color(red)"vertex")=-b/(2a)`

`y=-3x^2+5x+6" is in standard form"`

`"with " a=-3,b=5,c=6`

`rArrx_(color(red)"vertex")=-5/(-6)=5/6`

`"substitute this value into the function for y-coordinate"`

`rArry_(color(red)"vertex")=-3(5/6)^2+5(5/6)+6=97/12`

`rArrcolor(magenta)"vertex "=(5/6,97/12)`

Answer 3

`(5/6,97/12)`

Explanation:

`y=ax^2+bx+c` [Standard Form of a Quadratic Equation]
`y=-3x^2+5x+6`

`a = -3`
`b = 5`
`c = 6`

TO FIND THE X-VALUE OF THE VERTEX:
Use the formula for the axis of symmetry by substituting values for `b` and `a`:
`x = (-b)/(2a)`
`x = (-5)/(2(-3))`
`x = (-5)/-6`
`x = 5/6`

TO FIND THE Y-VALUE OF THE VERTEX:
Use the formula below by substituting values for `a`, `b`, and `c`:
`y = (-b^2)/(4a)+c`
`y = (-(5)^2)/(4(-3))+6`
`y = (-25)/(-12)+6`
`y = 25/12+72/12`
`y = 97/12`

Express as a coordinate.
`(5/6,97/12)`