Pls help !!!!!!!!!!!!!!

Mathematics

1 answer:

garri49 [273]3 years ago

6 0

The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola. For a quadratic function in standard form, y=ax2+bx+c , the axis of symmetry is a vertical line x=−b2a .

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If cos(θ)= -2/3 and θ lies in quadrant II, find the value of cotθ

Lera25 [3.4K]

$\text{Given that,}\\\\\cos \theta = - \dfrac 23\\\\\implies \cos^2 \theta = \dfrac 49\\\\\implies 1- \sin^2 \theta = \dfrac 49\\\\\implies \sin^2 \theta =1-\dfrac 49\\\\\implies \sin^2 \theta =\dfrac 59\\\\\\\text{Now,}\\\\\cot^2 \theta = \dfrac{\cos^2 \theta}{\sin^2 \theta} \\\\\\\implies \cot^2 \theta = \dfrac{\tfrac 49}{\tfrac 59}\\\\\\\implies \cot^2 \theta = \dfrac 49 \times \dfrac 95\\\\\\\implies \cot^2 \theta = \dfrac 45\\\\\\$

$\implies \cot \theta = \pm\sqrt{\dfrac 45} = \pm \dfrac 2{\sqrt 5}\\\\\\\text{Since}~ \theta ~ \text{lies in the second quadrant,} \cot \theta ~\text{will be negative.}\\\\\text{Hence,}~ \cot \theta = - \dfrac{2}{\sqrt 5}$