# What is the vertex of x – 4y^2 + 16y – 19 = 0?

Oct 22, 2017

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

#### Explanation:

This is a quadratic in $y$ instead of $x$. Thus the graph is like one done in $x$ but rotated ${90}^{0}$. So you would have an axis of symmetry that is parallel to the y-axis.

As the coefficient of ${y}^{2}$ is positive the graph is of form $\subset$

Write as $x = 4 {y}^{2} - 16 y + 19 \text{ } \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$

Now write as $x = 4 \left({y}^{2} - \frac{16}{4} x\right) + 19 \text{ } \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left({1}_{a}\right)$

Normally I would use the cheat of
${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

However in this case we have
${y}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(- \frac{16}{4}\right) = + 2$

Note that this is part of the process of completing the square.
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So by substitution for $y$

${x}_{\text{vertex}} = 4 {\left(+ 2\right)}^{2} - 16 \left(+ 2\right) + 19 = 3$

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