Question

If the parabola $y=a{x}^2-6x+b$ passes through $(0,2)$ and has its tangent at $x=\frac{3}{2}$ parallel to the $\mathrm{x}$-axis then

• A. $a=b=0$
• B. $a=b=1$
• C. $a=b=2$
• D. $a=b=-1$

Answer 1

The correct option is C

$a=b=2$

Explanation for the correct option.

Given an equation of a parabola $y=a{x}^2-6x+b$.
Since it passes through $(0,2)$ so it satisfies this point.
$$\begin{gathered} 2=0+b \\ b=2 \end{gathered}$$
Also, $\frac{dy}{dx}=2ax-6$
So, the slope of the tangent at $\mathrm{x}=\frac{3}{2}$
$\begin{array}{rcl}\frac{dy}{dx}& =& 2\mathrm{a}\left(\frac{3}{2}\right)-6\\ & =& 3\mathrm{a}-6\end{array}$
Since the tangent at $x=\frac{3}{2}$ is parallel to $x$-axis.
$$\begin{gathered} \Rightarrow 3a-6=0 \\ \Rightarrow a=2 \end{gathered}$$
So, $a=2,b=2$

Hence, option C is correct.