067basictpform

 Year 9 into 10 Step Up Program 11F

**Parabolas**
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Basic Turning Point Form
y = ±(x – h)² + k

Each part of the equation does a specific thing to the standard parabola ( y = x²)


 * A negative sign in front of the bracket reflects the parabola across the x-axis.
 * This means the parabola is __inverted__ (upside down).[[image:067grph1.gif width="316" height="258" align="right"]]


 * If there is no negative sign in front of the bracket, the parabola is __upright__.


 * The value inside the bracket with the x (h) shifts the parabola to the left or right in the __**opposite**__ direction to the sign.
 * The x-value of the turning point is h (reverse the sign).


 * The value after the bracket (+ k) shifts the parabola up or down in the **same** direction as the sign.
 * The y-value of the turning point is k

__Turning point__ is at (h, k)

__Axis of symmetry__ is a vertical line through the turning point : x = h

Find the __y-intercept__ by substituting x = 0 into the rule.

Find the __x-intercepts__ by substituting y = 0 and solving for x


 * Don't forget to label the axis with X and Y and label the graph with its equation.
 * Label the intercepts and the turning point with their coordinates.

** Example **
Sketch y = (x + 3)² – 2

__**Solution**__

Parabola is __upright__

Parabola shifted __left 3__ and __down 2__

Turning point : (–3, –2) Axis of symmetry : x = –3

y-intercept math \\ . \qquad y = (0+3)^2-2 \\ \\ . \qquad y = 7 \\ \\ . \qquad \text{coordinates : } (0, \; 7) math

x-intercepts math \\ . \qquad 0 = (x+3)^2 -2 \\ \\ . \qquad (x+3)^2 = 2 \\ \\ . \qquad x+3 = \pm1.414 \qquad \{ \textit{square root of 2} \} math

math \\ . \qquad x+3 = -1.4 \;\; \textit{ and } \;\; x+3 = 1.4 \\ \\ . \qquad x = -4.4 \quad \textit{ and } \quad x = -1.6 \\ \\ . \qquad \text{coordinates : } (-4.4, \; 0) \;\; \textit{ and } \;\; (-1.6, \; 0) math

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