Finding the values of the coefficients $a,b,c$ in the curve $y=ax^2+bx+c$.

Question

The curve $y=ax^2+bx+c$ crosses the $y$-axis at the point $(0,3)$ and has a stationary point at $(1,2)$. Find the value of $a$, $b$ and $c$.

Answer 1

HINT

By the givens we can easily find that

• for crossing $y$ at $(0,3) \implies 3=a\cdot 0^2+b\cdot 0+c$
• stationary point at $(1,2) \implies y'(1)=0\implies 2a\cdot 1+b=0$
• for crossing at $(1,2) \implies 2=a\cdot 1^2+b\cdot 1+c$

then solve the system of three equations to find $a,b, c$.

Answer 2

Let $f(x)=a x^2 + b x +c$. The value at $x=0$ give $c=f(0)$ and the stationary point is at $1=x_0=-\frac{b}{2a}$ with value $2=a x_0^2 +b x_0 +c$.

That give you as set of $3$ linear equations and $3$ unknowns.

Answer 3

@Gimusi, here... $y=ax^2+bx+c$ At point (0,3), We have $3=a•0^2+b(0)+c$ => 3=c

At stationary point (1,2)...we takes the derivative of "y"..and equate it to zero $=> y^1=2ax + b = 0$ There4,
$2ax + b = 0$ but x=1 $2a + b =0.....(1)$ And also, at point (1,2) We have, $2=a+b+3$ since c=3. $=>a+b=-1.......(2)$ $(1)-(2)$ :•a=1 Put a=1 into (2)... Then b=-2 Therefore, a=1,b=-2 and c=3. So, $y=x^2-2x+3$.

I'm a right?