Question

Find the vertex of the parabola described by the equation: $y={5(x-3)}^2-3$

• A. $(3,-3)$
• B. $(3,0)$
• C. $(0,-3)$
• D. $(-3,3)$

Answer 1

The correct option is A $(3,-3)$

Given equation of parabola is $y=5(x-3{)}^2-3$

$\Rightarrow y=5(x^2-6x+9)-3$

$\Rightarrow y=5x^2-30x+45-3$

$\Rightarrow y=5x^2-30x+42$

As per equation, here $a=5$ and $b=-30$

So $x$-coordinates of vertex of parabola is $=$ $-\frac{b}{2a}=-\frac{-30}{2\times 5}=\frac{30}{10}=3$

Put this value in equation, we get

$y=5(3-3)-3=-3$

Then vertex of the parabola is $(3,-3)$.