Finding constants of a given curve

Question

Find constants $a$, $b$ and $c$ such that the curve $y= ax^2 + bx +c$ passes through the point $(0,3)$ and has a relative extremum at $(1,2)$? I tried substituting the values of the given coordinates into the equation.

Answer 1

Your equation must satisfy

$$y(0)=3,\;y(1)=2\quad\text{and}\quad y'(1)=0$$

$$\begin{cases}3=a\cdot0+b\cdot0+c\\ 2=a\cdot1+b\cdot1+c\\ 0=2a\cdot1+b\end{cases}$$

$$\begin{cases}3=c\\ 2=a+b+c\\ 0=2a+b\end{cases}$$

$$\begin{cases}2=a+b+3\\ 0=2a+b\end{cases}$$

You get from here that $a+b=-1$ so $$0=a+a+b=a-1\implies {a=1\land b=-2}$$

So your curve has the following equation

$$y=x^2-2x+3$$

Answer 2

we have $$f(x)=ax^2+bx+c$$ and $$f(0)=c=3$$ thus we have $$f(1)=a+b+3=2$$ and it must be $$a+b=-1$$ for the extremum we get $$f'(x)=2ax+b$$ and we get two equations: $$a+b=-1$$ $$2a+b=0$$ can you proceed from here?

Answer 3

So I am not familiar with the term 'relative extremum' but given that its an equation of a parabola I think they mean that its vertex is at (1,2). If I am right then follow along :

1. (0,3) - we have our c = 3

2. Using the (1,2) we get : $2 = a(1)^2+b(1)+3$**

3. Solving for a we get : $a=-b-1$

4. We know that the vertex of a parabola is $\frac{-b}{2a}$

5. Now we substitute the a expression from point 3 in the vertex formula and we get $\frac{-b}{-2b-2} = \frac{b}{2b+2}$

6. $\frac{b}{2b+2} = 1$ which is equal to $b = -2$

7. $a = -b -1 \rightarrow a = 1$

8. So we get a=1 , b=-2, c=3

Let me know if i got the 'relative extremum' wrong lel

Answer 4

$3=a.0+b.0+c$ so $c=3$ for $(0,3)$ then similarly for $(1,2)$ so $2=a+b+3$ thus $a+b=-1$ and $f'(x)=0$ for extremum at $(1,2)$ hence $0=2a+b$ thus solving you get $a=1,b=-2$ thus the curve is a parabola with equation $x^2-2x+3=0$. Thats it

Answer 5

The extremum occurring at $(1,2)$ implies that the equation must be of the form

$$y=a(x-1)^2+2.$$

(This is a translated version of $y=ax^2$.)

Then plugging the known point,

$$3=a(0-1)^2+2\implies a=1$$and

$$y=x^2-2x+3.$$

Answer 6

Given c =3. Next, by differentiation of given parabola equation

$ -b/(2a) = 1, c - b^2/(4 a) = 2 $ , solve for $a,b.$