Question

1. $f(x) = -2(x + 5)^2 - 3$

Explanation:

Assuming the problem is to analyze the quadratic function \( f(x) = -2(x + 5)^2 - 3 \) (I think there might be a typo and it's \( -2(x + 5)^2 - 3 \) or maybe \( -2(x + 5)^2 + 3 \), but let's proceed with the given as \( f(x)=-2(x + 5)^2 - 3 \)) and maybe graph it or find its vertex, direction of opening etc.

Step1: Recall the vertex form of a quadratic function

The vertex form of a quadratic function is \( f(x)=a(x - h)^2 + k \), where \((h,k)\) is the vertex of the parabola and \(a\) determines the direction and the width of the parabola. If \(a>0\), the parabola opens upwards; if \(a < 0\), it opens downwards.

Step2: Identify \(a\), \(h\) and \(k\) from the given function

Given \( f(x)=-2(x + 5)^2 - 3 \), we can rewrite \( (x + 5) \) as \( (x-(-5)) \). So comparing with \( f(x)=a(x - h)^2 + k \), we have:

• \( a=-2 \)
• \( h=-5 \)
• \( k = - 3 \)

Step3: Determine the vertex and the direction of opening

• Vertex: The vertex of the parabola is at the point \((h,k)=(-5,-3)\)
• Direction of opening: Since \( a=-2<0 \), the parabola opens downwards.

Step4: (Optional) To graph the function, we can find a few more points. Let's find the value of \(f(x)\) when \(x=-4\):

Substitute \(x = - 4\) into \(f(x)\):
\( f(-4)=-2(-4 + 5)^2-3=-2(1)^2-3=-2 - 3=-5 \)
So the point \((-4,-5)\) is on the parabola.

When \(x=-6\):
\( f(-6)=-2(-6 + 5)^2-3=-2(-1)^2-3=-2 - 3=-5 \)
So the point \((-6,-5)\) is on the parabola.

Answer:

• Vertex: \((-5,-3)\)
• Direction of opening: Downwards
• Some points on the parabola: \((-4,-5)\), \((-6,-5)\) (and the vertex \((-5,-3)\))

If the original function was supposed to be \( f(x)=-2(x + 5)^2 + 3 \) (with a plus sign instead of minus before 3), then \(k = 3\) and the vertex would be \((-5,3)\) and the value at \(x=-4\) would be \(f(-4)=-2(1)^2 + 3=-2 + 3 = 1\), \(x=-6\) would give \(f(-6)=-2(-1)^2+3=-2 + 3 = 1\) and the direction of opening would still be downwards since \(a=-2<0\).