Question

If the points $(1,-8),(2,-1),(3,4)$ and $(5,8)$ lies on the graph of $f\left(x\right)=a{x}^2+bx+c$ and $f\left(x\right)$ is maximum at $x=5$, then point which must lie on the graph of $f\left(x\right)$ is

• A. $(4,6)$
• B. $(6,4)$
• C. $(8,-1)$
• D. $(10,-8)$

Answer 1

Answer: (C)

$(8,-1)$

Points $(1,-8),(2,-1),(3,4),(5,8)$ lie on the graph of $f\left(x\right)=a{x}^2+bx+c$

Also, $f\left(x\right)$ is maximum at $x=5$, implying $f^{\prime }\left(x\right)=0$ at $x=5$

$\Rightarrow -8=a+b+c$ ...(1)

$-1=4a+2b+c$ ...(2)

$4=9a+3b+c$ ..(3)

$8=25a+5b+c$ ...(4)

$0=2a\left(5\right)+b=10a+b$ ...(5)

Equation (5) gives $b=-10a$ ...(6)

Subtracting equations (1) and (2) gives

$3a+b=7$ ...(7)

Equations (6) and (7) would yield $-7a=7$ or $a=-1$

$\Rightarrow b=10$ and $c=-17$

It can be confirmed that equations (3) and (4) also satisfy these values.

Now, looking at the options, we substitute the $x$ coordinate and tally the answer with the given $y$ coordinate.

$f\left(4\right)=(-1)(4{)}^2+10(4)-17=7\ne 6$

$f\left(6\right)=(-1)(6{)}^2+10(6)-17=7\ne 4$

$f\left(8\right)=(-1)(8{)}^2+10(8)-17=-1$