What is the vertex of $y = 3 {(x + 5)}^{2} - 30 x - 25$ $y= 3(x+5)^2-30x-25$ ?

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Answer 1 · 2018-05-24T11:34:40

$50$ $50$

$y = 3 \cdot {(x + 5)}^{2} - 30 x - 25$ $y=3*(x+5)^2-30x-25$

= $3 \cdot (x^{2} + 10 x + 25) - 30 x - 25$ $3*(x^2+10x+25)-30x-25$

= $3 x^{2} + 30 x + 75 - 30 x - 25$ $3x^2+30x+75-30x-25$

= $3 x^{2} + 50$ $3x^2+50$

Thus, vertex of $y$ $y$ is $50$ $50$

What is the vertex of y=3(x+5)2−30x−25y= 3(x+5)^2-30x-25 ?