What is the minimum value of F(x)= 10x^{2} -6x-3 3/10 or -39/10

Mathematics

1 answer:

kolezko [41]3 years ago

6 0

Answer: 3/10

Step-by-step explanation:

Take the derivative of $10x^2-6x-3$ to get $20x-6$ . Set that equal to 0 to find the critical points of the function. The critical points is when the slope is either 0 or undefined.

Now do:

$20x-6=0\\20x=6\\x=\frac{6}{20} = \frac{3}{10}$

There are quite a few more steps to actually find the minimum, but for this example you can automatically assume its a minimum because it is the only critical point of the function. Ill show you these extra steps tho.

Plug in two numbers into the derivative. One that is less than 3/10 and one that is greater than 3/10. The numbers 0 and 1 are fine. When x = 0, the function is -6. When x = 1, the function is +14. A switch from negative to positive indicates a minimum value

You might be interested in

Helpppppppppppppppppppp

vagabundo [1.1K]

Answer:

the cyan one b is greater than or equal to -16

Step-by-step explanation:

okay so b is at least (at least being the keyword) -16, at least means that -16 is the minimum

hope this helps:)

4 0

2 years ago

PLEASE HELP ASAP

Mashutka [201]

Answer:

Step-by-step explanation:

3 0

2 years ago

Explain how you could write a quadratic function in factored form that would have a vertex with an x-coordinate of 3 and two dis

skad [1K]

The definition of quadratic function is given by:

 $(1) \ f(x)=ax^{2}+bx+c \\ \\ where \ a, b, c \ are \ real \ values$ 

So the problem asks for two conditions:

1. Write the quadratic function in factored form. 
2. The vertex has an x-coordinate of 3.

Then the equation (1) must be written as follows:

 $f(x)=(x-m)(x-n) \\ \\ where \ m \ and \ n \ are \ the \ roots$ 

To satisfy the two conditions:

$\frac{m+n}{2}=3$

So we are free to choose the value of one root, say:

$m=1$

Thus:

$n=6-1=5$

Finally, the answer is:

$f(x)=(x-1)(x-5)$

The graph of this function is shown in the figure below.