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What is the maximum value of -4z^2+20z-6?

Sagot :

caylus

Answer:

Hello,

19

Step-by-step explanation:

2 methodes:

1)

[tex]y=-4x^2+20x-6\\\\=-4(x^2-5x)-6\\\\=-4(x^2-2*\dfrac{5}{2}*x+\dfrac{25}{4} ) +25-6\\\\=-4(x-\frac{5}{2} )^2+19\\\\Maximum\ =19 \ if\ x=\dfrac{5}{2} \\[/tex]

2)

y'=-8x+20=0 ==> x=20/8=5/2

and y=-4*(5/2)²+20*5/2-6=-25+50-6=19

facundo3141592

finding maximums/minimums in quadratic equations:

The maximum value is 19

We want to find the maximum value of:

y = -4z^2+20z-6

Here, you can see that we have a quadratic equation with a negative leading coefficient.

This means that the arms of the graph will go downwards. From this, we can conclude that the maximum will the at the vertex (the highest point).

Remember that for a general equation like:

y = a*x^2 + b*x + c

The x-value of the vertex is:

x = -b/(2a)

(you can see that the variable is a different letter, that does not matter, is just notation)

Then for our equation:

y = -4z^2+20z-6

The z-value of the vertex is:

z = -20/(2*-4) = -20/-8 = 5/2

Then the maximum of the equation:

y = -4z^2+20z-6

is that equation evaluated in z = 5/2

So we get:

y = -4*(5/2)^2 + 20*(5/2) - 6  = 19

The maximum value is 19

If you want to learn more about this topic, you can read:

https://brainly.com/question/14336752

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