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Re: f(x) = 4x^2 + 28x + 49 for all x [#permalink]
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There is a equation to find the vertex of parabola \((x,y)\). A parabola is a u-shaped curve where the vertex is either the minimum point or the maximum point of the parabola, in this instance the vertex is the minimum point.
The method to find the value of \(y\) is a bit complicated however finding \(x\) is comparatively easier.
Here we just need to find the \(x\) value because at this point the value of \(y\) is also minimum.

the x point of the vertex of the parabola is \(\frac{-b}{2a}\) in the quadratic eqn of the form \(ax^2 + bx + c\) if a is +ve vertex is the minimum point else if a is -ve vertex is the maximum point
in the given problem \(4x^2 + 28x + 49\)
\(b = 28\) hence \(-b = -28\)
\(2a = 2*4 = 8\)
therefore \(x = -28/8 = -3.5\)
since \(-3.5<-3\)
option B
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Re: f(x) = 4x^2 + 28x + 49 for all x [#permalink]
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Re: f(x) = 4x^2 + 28x + 49 for all x [#permalink]
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