Chapter 9.1, Problem 79PFA

a.

To determine

To evaluate the domain of the function

Answer to Problem 79PFA

D. all real numbers

Explanation of Solution

Given information :

Quadratic function provided.

  $y=x^2+4x+3$

Options.

A. $\left\{x\right|x\ge 1\}$

B. $\left\{x\right|x\le -3\}$

C. $\left\{x\right|x\ge -1\}$

D. all real numbers

The graph represented by the function is that of a parabola.

The domain of a quadratic function is $\left\{-\infty <x<\infty \right\}$ , that is, all real numbers.

Option D. all real number; is the correct answer.

b.

To determine

To evaluate the range of the function

Answer to Problem 79PFA

A. $\left\{y\right|y\ge -1\}$

Explanation of Solution

Given information :

Quadratic function provided.

  $y=x^2+4x+3$

Options.

A. $\left\{y\right|y\ge -1\}$

B. $\left\{y\right|y\le -1\}$

C. $\left\{y\right|y\ge 1\}$

D. all real numbers

Formula used :

Formula to compute the x-coordinate of the vertex is $x=-\frac{b}{2\times a}$ . Here, $a=1$ and $b=4$ . ‘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.

  $x=-\frac{4}{2\times (1)}$

Putting the values of ‘a’ and ‘b’ .

  $x=-\frac{4}{2}$

Simplifying

  $x=-2$

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.

  $y={(-2)}^2+4(-2)+3$

Putting the value of $x=-2$ in the original function.

  $=4-8+3$

Simplifying the expression.

  $=-1$

Thus, the minimum is at $-1$ .

Since x can take up any real values, the domain is $\left\{-\infty <x<\infty \right\}$ .

Since the range is the values of y, it can only take up values greater than or equal to the minimum. That is, $\left\{y|y\ge -1\right\}$ .

Therefore, option A. $\left\{y\right|y\ge -1\}$ is the correct answer.

c.

To determine

To find the vertex of the function

Answer to Problem 79PFA

C. $(-2,-1)$

Explanation of Solution

Given information :

Quadratic function provided.

  $y=x^2+4x+3$

Options.

A. $(-4,3)$

B. $(-2,1)$

C. $(-2,-1)$

D. $(0,3)$

Formula used :

Formula to compute the x-coordinate of the vertex is $x=-\frac{b}{2\times a}$ . Here, $a=1$ and $b=4$ . ‘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.

  $x=-\frac{4}{2\times (1)}$

Putting the values of ‘a’ and ‘b’ .

  $x=-\frac{4}{2}$

Simplifying

  $x=-2$

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.

  $y={(-2)}^2+4(-2)+3$

Putting the value of $x=-2$ in the original function.

  $=4-8+3$

Simplifying the expression.

  $=-1$

Thus, the minimum is at $-1$ .

Hence the co-ordinates for the vertex are $(-2,-1)$

Therefore, option C. $(-2,-1)$ ; is the correct answer.

d.

To determine

To find the points that lie on the parabola

Answer to Problem 79PFA

C. $(-2,-1)$

Explanation of Solution

Given information :

Quadratic function provided.

  $y=x^2+4x+3$

Options.

A. $(-4,0)$

B. $(-3,0)$

C. $(-2,1)$

D. $(-1,0)$

E. $(0,3)$

F. $(-1,8)$

G. $(2,12)$

Formula used :

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. On putting the value of x -coordinate in the quadratic function if the result is equivalent to the corresponding y -coordinate, it can be ascertained that if a point lies on the parabola.

Calculation :

Option A. $(-4,0)$

Substituting $x=-4$ in the quadratic equation

  $y={(-4)}^2+4(-4)+3$

Putting the value of $x=-2$ in the original function.

  $=16-16+3$

Simplifying the expression.

  $=3$

Therefore, $(-4,3)$ lies on the parabola and $(-4,0)$ does not. Hence, option A. $(-4,0)$ is not selected.

The same method is followed for all the options to check if the point lies on the parabola.

Options:

B. $(-3,0)$

D. $(-1,0)$

E. $(0,3)$

These three points are found to be on the parabola by the method mentioned above.