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How do you find the vertex and the intercepts for #y = x^2 + 10x - 7#?

Answer 1

vertex (-5, -32)

y = x^2 + 10x - 7 x-coordinate of vertex: x = -b/(2a) = -10/2 = -5 y-coordinate of vertex: y(-5) = 25 - 50 - 7 = -32 To find x-intercepts, make y = 0 and solve the quadratic equation: #x^2 + 10x - 7 = 0# #D = d^2 = b^2 - 4ac = 100 + 28 = 128# --> #d = +- 8sqrt2# There are 2real roots: #x = -b/(2a) +- d/(2a) = -10/2 +- (8sqrt2)/2 = -5 +- 4sqrt2# #x1 = -5 + 4sqrt2# #x2 = -5 - 4sqrt2#

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Answer 2

Eva Adamson

To find the vertex of the quadratic function ( y = x^2 + 10x - 7 ), use the formula ( x = \frac{{-b}}{{2a}} ), where ( a ) is the coefficient of ( x^2 ) (which is 1), and ( b ) is the coefficient of ( x ) (which is 10). So, ( x = \frac{{-10}}{{2(1)}} = -5 ). Substitute ( x = -5 ) into the equation to find ( y ). Then, to find the intercepts, set ( y = 0 ) and solve for ( x ). These are the x-intercepts. To find the y-intercept, set ( x = 0 ) and solve for ( y ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.