Question

If, $x$ be real, then the minimum value of ${x^2} - 8x + 17$ .
A. $ - 1$
B. $0$
C. $1$
D. $2$

Answer 1

Hint: If we equate the given expression to a variable $y$ then, we will see that the equation obtained will be a parabolic equation. According to the quadrant in which the parabola is lying, we will find the coordinates of its vertex. Proceed the procedure till we get the value of both the coordinates. Hence, the minimum value is obtained.

Formula Used: Determine the vertex of the $x$-coordinate using formula $\dfrac{{ - b}}{{2a}}$ . Then, substitute that value in the given expression.

Complete step-by-step solution:
Suppose the given expression is $y = {x^2} - 8x + 17$ .
The parabola will open up because the coefficient of ${x^2}$ is positive.
As a result, the function will only return the minimum value, which is the $y$-coordinate of the vertex. We must first determine the vertex $x$-coordinate before determining its $y$-coordinate.
Formula to determine a vertex $x$-coordinate is $ = \dfrac{{ - b}}{{2a}}$ .
Replace $a$ with $1$ and $b$ with $ - 8$
$ \Rightarrow \dfrac{{ - b}}{{2a}} = \dfrac{{ - ( - 8)}}{{2 \times 1}}$
So, a vertex $x$-coordinate is $4$
In the provided function, change $x$ to $4$ and obtain the vertex $y$-coordinate.
$y = {(4)^2} - 8 \times 4 + 17$
$y = 1$
The bare minimum is $1$ .
Hence, the correct option is C.

Note: We can also use an alternative method. The ${x^2}$ term can be used to determine if our equation yields a maximum or minimum. The vertex point will be a minimum if the term ${x^2}$ is positive; if it is negative, the vertex will be a maximum. Use the given equation to find the bare minimum point, after confirming that we have one.