Week 10 – Precalc 11 – Caleb’s Blog

During Week 10, we learned a new perspective of the quadratic function. We can figure out many characteristics of the graph of a quadratic function such as the coordinates of the vertex, the domain, the range, the direction of the opening, the equation of the axis of symmetry, and the intercepts by utilizing its equation in its Standard Form. One example is Question 2 from “Assessing Your Understanding” on page 334:

$y = \frac{1}{2} (x + 3)^2 - 2$

When comparing this equation with the quadratic equation in its Standard Form, you can quickly find the coordinates of the vertex ( $p$ , $q$ )

$y = a(x -p)^2 + q$

From the example above, $p =-3$ and $q =-2$ ; therefore, the coordinates of the vertex are ( $-3$ , $-2$ ).

The ( $a$ ) coefficient before the brackets is a positive $\frac{1}{2}$ ; therefore, the parabola graph opens up.

The domain is as follows: $x$ ∈ $R$ . Also, since the graph is opens up, the vertex is a minimum point with a y-coordinate of -2.

So, that means the range is: $y \geq -2$ .

In general, the equation of the axis of symmetry is $x = p$ ; therefore, the equation should be $x =-3$ .

To find the y-intercept, substitute $x = 0$ in the original equation and solve for $y$ .

$y = \frac{1}{2} (x + 3)^2 - 2$

$y = \frac{1}{2} (0 + 3)^2 - 2$

$y = \frac{1}{2} (3)^2 - 2$

$y = \frac{9}{2} - 2$

$y = \frac{9}{2} - \frac{4}{2}$

$y = \frac{5}{2}$ or $2\frac{1}{2}$

So, the y-intercept is $2\frac{1}{2}$ .

Finally, to find the x-intercepts, substitute $y$ = 0 in the original equation and solve for x.

So, the x-intercepts are -1 and -5.

By looking at the completed graph, the vertex, and the x- and y-coordinates, the reversal can be done as well. In other words, the graph can be written back to its equation in factored form, General Form, or Standard Form.

Another characteristic we learned to solve was $a$ coefficient, which represents the “stretch” of the graph. The following example was taken from extra review questions during class:

If the vertex (-1, 2) and a second point (4, 7) are known, what would the quadratic function be?

$y = a(x -p)^2 + q$

Using this Standard Form, we insert the known values into this equation.

Now, we can apply the value of $a$ into the equation:

$y = \frac{1}{5} (x + 1)^2 + 2$

Week 10 – Precalc 11

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