what is the maximum value of y=-10x^2+50x+1

Question

what is the maximum value of y=-10x^2+50x+1 for graphing quadratics. 

Answer 1

The graph is an inverted parabola.  The maximum value occurs at the vertex.  

 

For y = Ax² + Bx + C, the x-coordinate of the vertex is -B/(2A).

 

So, for the given parabola, x-coordinate of vertex = -50/(2(-10)) = 2.5

 

Maximum value = y-coordinate of vertex

 

                        = -10(2.5)² + 50(2.5) + 1 = 63.5

 

Answer 2

You need to find the vertex of the parabola.

y = -10x² + 50x + 1    a = -10, b = 50, c = 1

x = -b/2a

x = -50/-20

x = 5/2

 

y = -10(5/2)² + 50(5/2) + 1

y = -62.5 + 125 + 1

y = 63.5

Answer 3

Hello Krystal,

 

There are two ways in which you can do this:

(Option 1) Using x = -b / (2a)

(Option 2) Putting into vertex form using 'completing the square'.

 

Likely, you will be using Option 1, but just in case your teacher is doing an exercise in the second method...

 

*Option 1

x = -50 / (2(-10)) = -50 / -20 = 5/2.

 

Plug 5/2 in for x to get y, which will be the max value:

y = -10(5/2)² + 50(5/2) +1

⇒ y = -10(25/4) + 125 + 1

⇒ y = -125/2 + 126

⇒ y = -62.5 + 126

⇒ y = 63.5

 

* Option 2

y = -10x² + 50x + 1

⇒ y = -10(x² -5x) +1

⇒ y = -10(x² -5x +[25/4]) +1 + [125/2]

⇒ y = -10(x - [5/2])² + 1 + 62.65

⇒ y = -10(x - [5/2])2 + 63.5

 

The 63.5 in this form is the max value.

 

Hope the above helps.  Should you have any questions about the above, please feel free to message me directly.

 

Kind regards,

Michael Ehlers