# How do you find the intercept and vertex of y= -3(x-1)²-1?

Aug 2, 2016

Vertex$\to \left(x , y\right) = \left(1 , - 1\right)$
${y}_{\text{intercept}} = - 4$

No ${x}_{\text{intercept}} \in \mathbb{R}$

#### Explanation:

$\textcolor{b l u e}{\text{Determine the vertex}}$

This is the Vertex Form of a quadratic equation so you can virtually directly read off the coordinates of the vertex.

$y = - 3 {\left(x \textcolor{red}{- 1}\right)}^{2} \textcolor{b l u e}{- 1}$

${x}_{\text{vertex}} = \left(- 1\right) \times \textcolor{red}{\left(- 1\right)} = + 1$
${y}_{\text{vertex}} = \textcolor{b l u e}{- 1}$

Vertex$\to \left(x , y\right) = \left(1 , - 1\right)$
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$\textcolor{b l u e}{\text{Determine the y intercept}}$

Set $x = 0$ giving

y_("intercept")=-3(0-1)^2-1" "=" "-4
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$\textcolor{b l u e}{\text{Determine the x intercept}}$

Set $y = 0$ giving

$0 = - 3 {\left(x - 1\right)}^{2} - 1 \text{ } \leftarrow$ add 1 to both sides

$\implies 1 = - 3 {\left(x - 1\right)}^{2} \text{ } \leftarrow$multiply both sides by (-1)

$\implies - 1 = + 3 {\left(x - 1\right)}^{2} \text{ } \leftarrow$divide both sides by 3

$\implies - \frac{1}{3} = {\left(x - 1\right)}^{2} \text{ } \leftarrow$ square root both sides

$\implies \pm \sqrt{- \frac{1}{3}} = x - 1 \text{ } \leftarrow$ add 1 to both sides

$x = 1 \pm \sqrt{- \frac{1}{3}} \text{ } \rightarrow x \in \mathbb{C}$

As the determinant is negative the curve does not cross the x-axis nor is the axis a tangent to the curve.

Thus the roots are in the number range of 'Complex Numbers'
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$\textcolor{b l u e}{\text{ The question lists intercepts as 'singular' thus the x-axis roots are not required}}$