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What is the vertex of the following quadratic function?
y = x² - 4x + 8

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\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{\textit{\small a}}{\downarrow }}{1}x^2\stackrel{\stackrel{\textit{\small b}}{\downarrow }}{-4}x\stackrel{\stackrel{\textit{\small c}}{\downarrow }}{+8} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)


\left(-\cfrac{ -4}{2(1)}~~~~ ,~~~~ 8-\cfrac{ (-4)^2}{4(1)}\right) \implies\left( - \cfrac{ -4 }{ 2 }~~,~~8 - \cfrac{ 16 }{ 4 } \right) \\\\\\ \left( \cfrac{ 4 }{ 2 }~~,~~8 - \cfrac{ 16 }{ 4 } \right)\implies \left( \cfrac{ 4 }{ 2 }~~,~~8 - 4 \right)\implies (~2~~,~~ 4~)

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