How do you find the vertex and the intercepts for $y - 10 = 15 + x^2 + 2x$ ?

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Answer 1 · 2017-02-07T05:29:19

minimum at $(-1,24)$ and y-intercept is $25$

$y-10 = 15+x^2+2x$

rearrange,
$y = x^2+2x+25$

y - intercept, when x=0,
$y = 25$

there is no x-intercept since $b^2-4ac < 0$ where $a=1, b=2 and c=25$

by completing a square,
$y=(x+1)^2-(1)^2+25$
$y=(x+1)^2+24$
This is a minimum vertex because there is a +ve coefficient for $(x+1)^2$

therefore, it vertex is minimum at $(-1,24)$ and y-intercept is $25$

How do you find the vertex and the intercepts for y - 10 = 15 + x^2 + 2xy - 10 = 15 + x^2 + 2x?