Chapter SH, Problem 9.5EP

To determine

To find: the equation in standard form and identify the vertex, axis of symmetry and direction of opening of the parabola.

Answer to Problem 9.5EP

At $\text{x=-3}$ and $\text{y=-2}$ are the vertex of the equation, direction of the parabola will open upward and axis of symmetry is at $\text{x=-3}$ .

Explanation of Solution

Given:

  ${\text{y=x}}^{\text{2}}\text{+6x+7}$ .

Concept used:

Vertex Is lowest point if parabola open upward.

Similarly, vertex is maximum point if parabola open downward.

The axis of symmetry is a line that reflects the parabola around, and its symmetric.

If the sign of coefficient of the ${\text{x}}^{\text{2}}$ is positive which means parabola is open upward and if its negative then parabola is open downward.

Vertex of the quadratic equation $\text{=}\left(\frac{\text{-b}}{\text{2a}}\text{,f}\left(\frac{\text{-b}}{\text{2a}}\right)\right)$ .

Calculation:

  ${\text{y=x}}^{\text{2}}\text{+6x+7}$ .

Here the direction of the parabola will open upward since the sign of coefficient of the ${\text{x}}^{\text{2}}$ is positive.

Therefore, the point or vertex will be minimum.

The axis of symmetry can measure as:

Compare the given Equation:

  ${\text{y=x}}^{\text{2}}\text{+6x+7}$ and ${\text{y=ax}}^{\text{2}}\text{+bx+c}$ . Where $\text{a}\ne \text{0}$ .

So, the symmetry will be at $\text{x=-}\frac{\text{b}}{\text{2a}}$ .

  $\Rightarrow \text{x=-}\frac{6}{2\left(1\right)}$ .

  $\Rightarrow \text{x=-}\frac{6}{2\left(1\right)}$ .

  $\text{x=-3}$ .

The axis symmetry of the given equation will be at $\text{x=-3}$ .

Vertex of the parabola can be solved as:

  $\left(1\right)$ .

By completing the square:

  ${\text{y=x}}^{\text{2}}\text{+2}\left(\text{x}\right)\left(\text{3}\right){\text{+3}}^{\text{2}}{\text{-3}}^{\text{2}}\text{+7}$ .

  $\text{y=}{\left(\text{x+3}\right)}^2{\text{-3}}^{\text{2}}\text{+7}$ .

  $\text{y=}{\left(\text{x+3}\right)}^2\text{-2}$ .

Here at $\text{x=-3}$ and $\text{y=-2}$ are the vertex of the equation.

Or

  $\left(2\right)$ .

Vertex of the quadratic equation $\text{=}\left(\frac{\text{-b}}{\text{2a}}\text{,f}\left(\frac{\text{-b}}{\text{2a}}\right)\right)$ .

  ${\text{y=x}}^{\text{2}}\text{+6x+7}$ .

Here $\text{a=1, b=6 and c=7}$ .

Putting in vertex of the equation.

  $\left(\frac{\text{-b}}{\text{2a}}\text{,f}\left(\frac{\text{-b}}{\text{2a}}\right)\right)$ .

  $\Rightarrow \left(\frac{\text{-6}}{\text{2}\left(1\right)}\text{,f}\left(\frac{\text{-6}}{\text{2}\left(1\right)}\right)\right)$ .

  $\Rightarrow \left(-3\text{,}-2\right)$

Hence, at $\text{x=-3}$ and $\text{y=-2}$ are the vertex of the equation, direction of the parabola will open upward and axis of symmetry is at $\text{x=-3}$ .