1. completing the square

If the function ax^2+bx+c has a minimum value -5 when x=-1 and 0 when x=-2 , find the values of a , b and c .

This is what i did :

Let $f(x)=ax^2+bx+c$.. completing the square

$=a[(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a^2}]$

$=a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}$

$x=-1$ when $y=-5$

$x=-\frac{b}{2a}$ ... b=2a -- 1

$\frac{4ac-b^2}{4a}=-5$ ... 4ac-b^2=-20a -- 2

Combining gives me $ac-a^2=5a$ --3

x=-2 when y=0

$-2=-\frac{b}{2a}$--- 4a=b ... 4

$4ac-b^2=0$ ... 5

combining , $ac-4a^2=0$ --- 6

Now , combining 3 and 6 , $3a^2-5a=0$ .. $a=0$ or $a=\frac{5}{3}$

i just couldnt find my mistake . thanks .

2. line of symmetry is x = -b / 2a = -1, so b = 2a.
x = -2 is a zero and just 1 left of line of symmetry, so other zero must be 1 right of line of symmetry at x = 0.
finally, (-2,0) is 1 left but 5 up from min (-1,-5), so vertical stretch is 5 = a.

so 5x(x + 2) = 5x² + 10x, a = 5, b = 10, c = 0

3. $f(-2)=0\Rightarrow 4a-2b+c=0$

(When x=-2 the value of is not minimum)

4. $f\left( { - 2} \right) = 4a - 2b + c = 0.$

$f{\left( { - 1} \right)_{\min }} = a - b + c = - 5 \Rightarrow f'\left( { - 1} \right) = - 2a + b = 0.$

Then we have a system of linear equations:

$\left\{ \begin{gathered}
4a - 2b + c = 0, \hfill \\
a - b + c = - 5, \hfill \\
- 2a + b = 0. \hfill \\
\end{gathered} \right.$

5. Good gosh, were you required to solve the problem using complete the square? You can see that the above solutions are much easier. Anyway, I found one error in your steps:

Originally Posted by thereddevils
If the function ax^2+bx+c has a minimum value -5 when x=-1 and 0 when x=-2 , find the values of a , b and c .

This is what i did :

Let $f(x)=ax^2+bx+c$.. completing the square

$=a[(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a^2}]$

$=a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}$

$x=-1$ when $y=-5$

$x=-\frac{b}{2a}$ ... b=2a -- 1

$\frac{4ac-b^2}{4a}=-5$ ... 4ac-b^2=-20a -- 2

Combining gives me $ac-a^2=5a$ --3
That should be $ac-a^2={\color{red}-}5a$. I didn't look further I got tired.

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6. Originally Posted by yeongil
Good gosh, were you required to solve the problem using complete the square?
I offer my solution for testing

7. Originally Posted by DeMath
Originally Posted by yeongil
Good gosh, were you required to solve the problem using complete the square?
I offer my solution for testing
Oops, my comment wasn't directed towards you, DeMath, but to thereddevils.

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