Answer:
See attached for graph of the given function.
Step-by-step explanation:
Vertex form of a quadratic function
[tex]f(x)=a(x-h)^2+k[/tex]
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:
[tex]g(x)=-\dfrac{1}{5}(x+5)^2-2[/tex]
Vertex
Comparing the given function with the vertex formula:
[tex]\implies h=-5[/tex]
[tex]\implies k=-2[/tex]
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:
[tex]\implies f(0)=-\dfrac{1}{5}(0+5)^2-2=-7[/tex]
Therefore, the y-intercept is (0, -7).
x-intercepts
To find the x-intercepts, set the function to zero and solve for x:
[tex]\implies -\dfrac{1}{5}(x+5)^2-2=0[/tex]
[tex]\implies -\dfrac{1}{5}(x+5)^2=2[/tex]
[tex]\implies (x+5)^2=-10[/tex]
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:
[tex]\implies f(-10)=-\dfrac{1}{5}(-10+5)^2-2=-7[/tex]
[tex]\implies f(5)=-\dfrac{1}{5}(5+5)^2-2=-22[/tex]
[tex]\implies f(-15)=-\dfrac{1}{5}(-15+5)^2-2=-22[/tex]
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.
