Chapter 9.1, Problem 80PFA

To determine

To evaluate the value of c

Answer to Problem 80PFA

For the value $c=-6$ in $y=x^2-2x+c$ , the range of the function is $\left\{y\right|y\ge -7\}$ .

Explanation of Solution

Given information :

Quadratic function provided.

  $y=x^2-2x+c$

The range is $\left\{y\right|y\ge -7\}$

Formula used :

Formula to compute the x-coordinate of the vertex is $x=-\frac{b}{2\times a}$ . Here, $a=1$ and $b=-2$ . ‘a’ and ‘b’ are coefficients $x^2$ and $x$ respectively. The corresponding value of ‘y’ for the vertex would be the maximum or minimum.

Calculation :

Since the coefficient of ‘a’ is positive, the curve of this function is upward facing and hence the function has a minimum value.

x -coordinate of the vertex is

  $x=-\frac{b}{2\times a}$

Formula for axis of symmetry.

  $x=-\frac{(-2)}{2\times (1)}$

Putting the values of ‘a’ and ‘b’ .

  $x=\frac{2}{2}$

Simplifying

  $x=1$

Vertex can be found out by putting the value of x computed in the axis of symmetry in the original function. This will give a value of y , that will be the minimum.

Here, the range of y is known to us, $\left\{y\right|y\ge -7\}$ . The minimum value that y can have is $y=-7$ .

Hence, the vertex is at $(1,-7)$

  $y=x^2-2x+c$

Putting the value of $x=1$ and $y=-7$ in the original function,

  $-7=1^2-2\times 1+c$

Simplifying the expression.

  $c=-6$

Thus, for the value $c=-6$ in $y=x^2-2x+c$ , the range of the function is $\left\{y\right|y\ge -7\}$ .