How can you find the coordinates of the turning point of a quadratic graph?

To find the coordinates of the turning point of a quadratic graph, you can either: complete the square differentiate and set the derivative equal to zero to find the x -coordinate

IGCSEMathsCambridge (CIE)ExtendedFlashcards

Quadratic Graphs (Cambridge (CIE) IGCSE Maths)

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FrontQuadratic Graphs
Will the vertex of the graph $y = - x^{2} + x + 5$ be a maximum or a minimum point?
FrontQuadratic Graphs
Where does the graph of $y = 3 (x - 1) (x + 2)$ cross the x-axis?
BackQuadratic Graphs
The graph of $y = 3 (x - 1) (x + 2)$ crosses the x-axis at $(1, 0)$ and $(- 2, 0)$ .
To find these roots, make each factor equal to zero to find the x-coordinate:
$x - 1 = 0$ gives $x = 1$
$x + 2 = 0$ gives $x = - 2$
FrontQuadratic Graphs
How can you find the coordinates of the turning point of a quadratic graph?
BackQuadratic Graphs
To find the coordinates of the turning point of a quadratic graph, you can either:
complete the square
differentiate and set the derivative equal to zero to find the x-coordinate

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Cards in this collection (4)

Will the vertex of the graph $y = - x^{2} + x + 5$ be a maximum or a minimum point?
The vertex of the graph $y = - x^{2} + x + 5$ will be a maximum point.
The vertex of a quadratic graph will be a maximum point if the coefficient of $x^{2}$ is negative. The graph is n-shaped.
Where does the graph of $y = 3 (x - 1) (x + 2)$ cross the x-axis?
The graph of $y = 3 (x - 1) (x + 2)$ crosses the x-axis at $(1, 0)$ and $(- 2, 0)$ .
To find these roots, make each factor equal to zero to find the x-coordinate:
- $x - 1 = 0$ gives $x = 1$
- $x + 2 = 0$ gives $x = - 2$
How can you find the coordinates of the turning point of a quadratic graph?
To find the coordinates of the turning point of a quadratic graph, you can either:
- complete the square
- differentiate and set the derivative equal to zero to find the x-coordinate
True or False?
The turning point of $y = 4 - {(x - 1)}^{2}$ is at the point $(4, 1)$ .
False.
The turning point of $y = 4 - {(x - 1)}^{2}$ is not at the point $(4, 1)$ .
The x-coordinate is the value that makes the squared bracket equal to zero.
The coordinates should be $(1, 4)$ .

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