How do you find the vertex and intercepts for (x+3)^2=16(y-2)?

Aug 20, 2017

Vertex $\to \left(x , y\right) = \left(- 3 , 2\right)$

y-intercpt $\to \left(0 , \frac{41}{16}\right)$

x-intercept $\to$ None

Explanation:

This is almost in 'completed square' form.

Sometimes called vertex form.

We need to get the $y$ on its own.

$y - 2 = \frac{1}{16} {\left(x + 3\right)}^{2}$

$y = \frac{1}{16} {\left(x \textcolor{red}{+ 3}\right)}^{2} \textcolor{b l u e}{+ 2} \leftarrow \text{ Now it is in vertex form}$
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$\textcolor{g r e e n}{\text{To determine the vertex}}$

${x}_{\text{vertex}} = \left(- 1\right) \times \left(\textcolor{red}{+ 3}\right) = - 3$

${y}_{\text{vertex}} = \textcolor{b l u e}{+ 2}$

Vertex $\to \left(x , y\right) = \left(- 3 , 2\right)$
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$\textcolor{g r e e n}{\text{To determine the " y" intercept}}$

This occurs at $x = 0$ so by substitution:

$y = \frac{1}{16} {\left(x + 3\right)}^{2} + 2 \text{ "->" } y = \frac{9}{16} + 2 = \frac{41}{16}$

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$\textcolor{g r e e n}{\text{To determine the " x" intercept}}$

This occurs at $y = 0$ so by substitution:

$y = \frac{1}{16} {\left(x + 3\right)}^{2} + 2 \text{ "->" } - 2 = \frac{1}{16} {\left(x + 3\right)}^{2}$

color(white)("mmmmmmmmmmmb")->" "-32=(x+3)^2

Square root both sides

$\sqrt{- 32} = x + 3 \textcolor{red}{\leftarrow \text{ This result has a problem}}$

You can take the square root of a negative number but the answer is in the realm of Complex Numbers. Consequently the graph does not cross the x-axis

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Foot Note: The ${x}^{2}$ term is positive so the graph is of shape $\cup$.

The vertex (which is a minimum value in this case) is above the x-axis. That alone tells you that there is no solution to $y = 0$ that is in the set of 'Real Numbers'. That is $y \ne 0 \text{ for } x \in \mathbb{R}$