Question

Help please! anyone know how to graph this

Answer 1

Answer:

The grpah should help!

Answer 2

Answer:

See attached for graph of the given function.

Step-by-step 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:

$g(x)=-\dfrac{1}{5}(x+5)^2-2$

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:

$\implies f(0)=-\dfrac{1}{5}(0+5)^2-2=-7$

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

x-intercepts

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

$\implies -\dfrac{1}{5}(x+5)^2-2=0$

$\implies -\dfrac{1}{5}(x+5)^2=2$

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

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:

$\implies f(-10)=-\dfrac{1}{5}(-10+5)^2-2=-7$

$\implies f(5)=-\dfrac{1}{5}(5+5)^2-2=-22$

$\implies f(-15)=-\dfrac{1}{5}(-15+5)^2-2=-22$

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.