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Help please! anyone know how to graph this

Help please! anyone know how to graph this-example-1
User Outluch
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4 votes

Answer:

The grpah should help!

Help please! anyone know how to graph this-example-1
User Likan Zhan
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8.5k points
3 votes

Answer:

See attached for graph of the given function.

Explanation:

Vertex form of a quadratic function


f(x)=a(x-h)^2+k

where:

  • (h, k) is the vertex.
  • a is some constant to be found.
    If a>0 the parabola opens upwards.
    If a<0 the parabola opens downwards.

Given function:


g(x)=-(1)/(5)(x+5)^2-2

Vertex

Comparing the given function with the vertex formula:


\implies h=-5


\implies k=-2

Therefore, the vertex of the parabola is (-5, -2).

As a<0, the parabola opens downwards. Therefore, the vertex is the maximum point of the curve.

Axis of symmetry

The axis of symmetry is the x-value of the vertex.

Therefore, the axis of symmetry is x = -5.

y-intercept

To find the y-intercept, substitute x = 0 into the given function:


\implies f(0)=-(1)/(5)(0+5)^2-2=-7

Therefore, the y-intercept is (0, -7).

x-intercepts

To find the x-intercepts, set the function to zero and solve for x:


\implies -(1)/(5)(x+5)^2-2=0


\implies -(1)/(5)(x+5)^2=2


\implies (x+5)^2=-10

As we cannot square root a negative number, the curve does not intercept the x-axis.

Additional points on the curve

As the axis of symmetry is x = -5 and the y-intercept is (0, -7), this means that substituting values of x in multiples of 5 either side of the axis of symmetry will yield integers:


\implies f(-10)=-(1)/(5)(-10+5)^2-2=-7


\implies f(5)=-(1)/(5)(5+5)^2-2=-22


\implies f(-15)=-(1)/(5)(-15+5)^2-2=-22

Therefore, plot:

  • vertex = (-5, -2)
  • y-intercept = (0, -7)
  • points on the curve = (-10, -7), (5, -22) and (-15, -22)
  • axis of symmetry: x = -5

Draw a smooth curve through the points, using the axis of symmetry to ensure the parabola is symmetrical.

Help please! anyone know how to graph this-example-1
User SSAMEERR
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8.5k points

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