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Convert to the standard form of the quadratic function
Answer
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Find the maximum and minimum of the quadratic function
Answer
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Calculate the differentiation
Answer
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Graph
$y = x ^ { 2 } - 3 x + 1$
$x$Intercept
$\left ( \dfrac { 3 } { 2 } - \dfrac { \sqrt{ 5 } } { 2 } , 0 \right )$, $\left ( \dfrac { \sqrt{ 5 } } { 2 } + \dfrac { 3 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , 1 \right )$
Minimum
$\left ( \dfrac { 3 } { 2 } , - \dfrac { 5 } { 4 } \right )$
Standard form
$y = \left ( x - \dfrac { 3 } { 2 } \right ) ^ { 2 } - \dfrac { 5 } { 4 }$
$y = x ^{ 2 } -3x+1$
$y = \left ( x - \dfrac { 3 } { 2 } \right ) ^ { 2 } - \dfrac { 5 } { 4 }$
Rewrite it as the standard form of the quadratic function
$y = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Add and subtract constants to convert the quadratic equation on the right side to the standard form $ $
$y = x ^ { 2 } - 3 x + 1 + \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$y = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } + 1 \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } - \left ( \dfrac { 3 } { 2 } \right ) ^ { 2 }$
$ $ Organize the expression using $ A^{2} ± 2AB + B^2 = (A ± B)^{2}$
$y = \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } + 1 - \left ( \dfrac { 3 } { 2 } \right ) ^ { 2 }$
$y = \left ( x - \dfrac { 3 } { 2 } \right ) ^ { 2 } + 1 - \left ( \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$ $ Calculate power $ $
$y = \left ( x - \dfrac { 3 } { 2 } \right ) ^ { 2 } + 1 - \color{#FF6800}{ \dfrac { 9 } { 4 } }$
$y = \left ( x - \dfrac { 3 } { 2 } \right ) ^ { 2 } + \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 4 } }$
$ $ Subtract $ \dfrac { 9 } { 4 } $ from $ 1$
$y = \left ( x - \dfrac { 3 } { 2 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } }$
$- \dfrac { 5 } { 4 }$
Find the maximum and minimum of the quadratic function
$\color{#FF6800}{ y } = \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$ $ Rewrite it as the standard form of the quadratic function $ $
$\color{#FF6800}{ y } = \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } }$
$\color{#FF6800}{ y } = \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } }$
$ $ As $ a \gt 0 $ is, the minimum value is $ - \dfrac { 5 } { 4 } $ if $ x = \dfrac { 3 } { 2 }$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } }$
$\dfrac {d } {d x } {\left( y \right)} = 2 x - 3$
Calculate the differentiation of the logarithmic function
$\dfrac {d } {d \color{#FF6800}{ x } } {\left( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right)}$
$ $ Calculate the differentiation $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
$ $ 그래프 보기 $ $
Quadratic function
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