Variation in graph with function f(x) = x^2 + kx +lposition (o,o) with parabola graph.Discribe the position of this graph f(x+c)

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The given equation of parabola is given in standard form.

Since the leading coefficient is >0, the parabola opens upwards.

The axis of symmetry of given parabola is parallel to y axis and it is given by:

x = -k/2*1

x = -k/2

We'll write the quadratic function f(x+c) = (x+c)^2 + k*(x+c) + l

f(x+c) = (x+c)^2 + k*(x+c) + l

f(x+c) = x^2 + 2cx + c^2 + kx + kc + l

f(x+c) = x^2 + x(2c + k) + c^2 + kc + l

Again, the leading coefficient is >0, therefore the parabola opens upwards.

The axis of symmetry of given parabola is parallel to y axis and it is given by:

x = -(2c + k)/2

x = -2c/2 - k/2

x = -k/2 - c

We'll consider the example:

f(x) = x^2 + 6x + 7

x = -6/2

x = -3

f(x+2) = x^2 + x(2*2 + 6) + 2^2 + 6*2 + 7

f(x+2) = x^2 + 10x + 23

x = -10/2

x = -5

We can see that the vertex of parabola was moved to the right by the amount of c = 2.

We notice that the green parabola is represented by the function f(x) = x^2 + 6x + 7, having the x coordinate of vertex, x = -3 and the orangle parabola is represented by the function f(x) = x^2 + 10x + 23, having the x coordinate of vertex, x = -5.

The axis of symmetry is displaced by the amount -c, therefore the vertex of parabola is moved to the left, with the amount -c.

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