Finding Maximum Values: Using Calculus to Solve Quadratic Equations

[ihopeican]

Question:
Hello, the question is to find the MAXIMUM VALUE of -3x^2-5x+1=0.

Attempt:

I tried to enter this into my calculate in Polynomial and got two values of x:
x=-1.85
and
x=0.1804

Thankyou!

See following post for attempted answer.

 

[ihopeican]

Question:
Hello, the question is to find the MAXIMUM VALUE of -3x^2-5x+1=0.

Attempt:

I tried to enter this into my calculate in Polynomial and got two values of x:
x=-1.85
and
x=0.1804

Thankyou!

I came up with -0.8333333333.

I got this from thinking the maximum value is just the turning point (Ie: -b/2a)

 

[colby2152]

Use the quadratic formula...

 

[HallsofIvy]

Using the quadratic formula to solve that equation will tell you where the value of y is 0. The x-value for maximum y will be halfway between the two roots.

By the way, you really want to find the maximum value of y= -3x^2-5x+1. An equation, like -3x^2-5x+1=0, doesn't have a "value" to begin with!

You can also find the maximum by completing the square in -3x^2-5x+1.

 

[colby2152]

Hmm, read it wrong due to the zero... I agree that the equation doesn't have a value to begin with.

 

[jaime2000]

I used calculus to get the maximum, and I get what ihopeican got; x = -0.8333... without calculus, you can get the x value of the midpoint by -b/2a, as ihopeican suggested, or by finding the zeros of the equation and finding the average of their x values, as HallsofIvy suggested. When you get to calculus, you will learn how to use derivatives to get maximums and minimums of lots more equations, not just quadratics. It's fun, easy, and fast. XD