# QUADRATICS questions !!!please answer questions 1, 2, 3 http://www.geogebra.org/en/upload/files/MSP/EvonnePankowski/Multimedia1_EHP.html Thank you

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You need to find how to relate the coefficients of quadratic to the shape of parabola.

You should know that if the coefficient a>0, then the parabola is concave up and if a<0, then the parabola is concave down.

You also need to remember that the x coordinate of vertex expresses the horizontal shift of parabola, while y coordinate represents the vertical shift.

You need to solve the equation `ax^2 + bx + c = 0` for x to find the horizontal shift, hence you should complete the square such that:

`a(x^2 + (b/a)x + c/a) = 0`

`(x^2 + 2*b/(2a)*x + b^2/(4a)) - b^2/(4a) + c/a = 0`

`(x+ b/(2a))^2 = b^2/(4a)- c/a`

You need to bring to a common denominator the terms from the right side such that:

`(x+ b/(2a))^2 = (b^2 - 4ac)/(4a)`

You need to take square root such that:

`x+ b/(2a) = +-sqrt((b^2 - 4ac)/(4a))`

`x = -b/(2a) +- (sqrt(b^2 - 4ac))/(2a)`

Hence, if `(b^2 - 4ac) = 0 =gt x = -b/(2a)`

The y coordinate of the vertex is `y = -Delta/(4a)` and this represents the vertical shift of parabola.

**Hence, the coefficient a indicates if the parabola is concave up or down and the coordinatesof the vertex indicate the horizontal and vertical shifts of parabola.**