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What is the vertex of f(x) = x2 + 10x + 16? A. (2, 6) B. (1, 5) C. (5, 9) D. (-5, -9)
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first coordinate of the vertex is always \(-\frac{b}{2a}\) which in your case is \(-\frac{10}{2}=-5\)
the vertex is the highest, or lowest point of a parabola it is expressed as -b/2a recognize this as part of the quadratic formula? the +/-sqrt(b^2-4ac) give the values on either side of the 'Line of Symmetry' so -b/2a= -10/2= -5 this is the x value plug -5 into the equation to get the value for y (-5)^2+10(-5)+16 25-50+16 -9 the vertex is (-5,-9)
is that right ?
second coordinate is what you get when you replace \(x\) by \(-5\) but since you only have one correct choice you don't even have to do that
yes
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