Quadratic funtion

[renii]

Could you please help me to answer this question

:
please show me the step by step process in answering this question.


The Quadratic function which takes the value 41 at x=-2 and the value 20 at x = 5 and is minimized at x = 2 is

y = __x² - __x +__

the minimum value of this function is ____.



---Thank you in advance---

 

[Unknown008]

I would put a, b and c in the blanks to make it easier to solve:

\(\displaystyle y = ax^2 + bx+ c\)

When x = -2, y = 41:

\(\displaystyle 41 = a(-2)^2 + b(-2)+ c\)

When x = 5, y = 20:

\(\displaystyle 20 = a(5)^2 + b(5)+ c\)

The equation is minimized at c, so that \(\displaystyle c = 2\)

Can you proceed from there?

EDIT: It is not minimised at c = 2. Assuming that you don't know how to take derivatives yet, you can turn the equation into its completed square form:

\(\displaystyle y = a\left(x+\dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a} + c\)

From there, you know that \(\displaystyle - \dfrac{b^2}{4a} + c\) is your minimum, which you can equate to 2.

 

[HallsofIvy]

I would put a, b and c in the blanks to make it easier to solve:

\(\displaystyle y = ax^2 + bx+ c\)

When x = -2, y = 41:

\(\displaystyle 41 = a(-2)^2 + b(-2)+ c\)

When x = 5, y = 20:

\(\displaystyle 20 = a(5)^2 + b(5)+ c\)

Yes, that is true.

The equation is minimized at c, so that \(\displaystyle c = 2\)

No, that is not true. You can use the "minimized" condition in either of two ways:
1) Complete the square. \(\displaystyle ax^2+ bx+ c= a(x^2+ (b/a)x+ b^2/4a^2) + c- b^2/4a= (x+ b/2a)^2+ c- b^2/4a\)
which is minimized at x= -b/2a= 2.
2) Differentiate. The minimum occurs where f'= 2ax+ b= 0 which again gives x= -b/2a= 2.

Can you proceed from there?

You now have the three equations, 4a- 2b+ c= 41, 25a+ 5b+ c= 20, and b=-4a. I recommend subtracting the first equation from the second to eliminate c, then using that result with b= -4a.

 

[Unknown008]

Yes, that is true.


No, that is not true. You can use the "minimized" condition in either of two ways:
1) Complete the square. \(\displaystyle ax^2+ bx+ c= a(x^2+ (b/a)x+ b^2/4a^2) + c- b^2/4a= (x+ b/2a)^2+ c- b^2/4a\)
which is minimized at x= -b/2a= 2.
2) Differentiate. The minimum occurs where f'= 2ax+ b= 0 which again gives x= -b/2a= 2.

Oops, sorry, you're right. I started completing the square but somehow I thought it gave the same result, while it almost never does.

@Denis: Been busy with stuff and remembered about here some hours ago.

 

[lookagain]

1) Complete the square.

\(\displaystyle ax^2 + bx + c = \)

\(\displaystyle a[x^2 + (b/a)x + b^2/(4a^2)] + c - b^2/(4a^2) =\)


\(\displaystyle a[x + b/(2a)]^2 + c - b^2/(4a^2)\)


which is minimized at x = -b/(2a) = 2.


2) Differentiate. The minimum occurs where f'= 2ax + b = 0 which again gives x = -b/(2a) = 2.

Careful! Don't forget the needed grouping symbols, any missing coefficients, and any missing exponents.