Question

1. $f(x)=x^{2}+6x + 5$ a b c vertex $x=-\frac{b}{2a}$ 6) $f(x)=-3x^{2}+6x + 1$ a b c vertex $x=-\frac{b}{2a}$ 7) $f(x)=3(x + 2)(x+6)$ x int: $x=\frac{p + q}{2}$ vertex: 8) $f(x)=0.5(x - 2)(x + 6)$ x int: $x=\frac{p + q}{2}$ vertex: 9) $f(x)=-x(x + 6)$ x int: $x=\frac{p + q}{2}$ vertex: -2

Explanation:

Step1: Recall vertex - formula for quadratic functions

For a quadratic function in the form \(y = ax^{2}+bx + c\), the \(x\) - coordinate of the vertex is \(x=-\frac{b}{2a}\). For a quadratic function in factored form \(y=a(x - p)(x - q)\), the \(x\) - coordinate of the vertex is \(x=\frac{p + q}{2}\), and then we substitute the \(x\) - value into the function to find the \(y\) - coordinate of the vertex.

Problem 5: \(f(x)=x^{2}+6x + 5\)

Here \(a = 1\), \(b = 6\), \(c = 5\).

Step2: Calculate the \(x\) - coordinate of the vertex

Using \(x=-\frac{b}{2a}\), we substitute \(a = 1\) and \(b = 6\):
\(x=-\frac{6}{2\times1}=-3\)

Step3: Calculate the \(y\) - coordinate of the vertex

Substitute \(x=-3\) into \(f(x)=x^{2}+6x + 5\):
\(f(-3)=(-3)^{2}+6\times(-3)+5=9-18 + 5=-4\)
The vertex is \((-3,-4)\)

Problem 6: \(f(x)=-3x^{2}+6x + 1\)

Here \(a=-3\), \(b = 6\), \(c = 1\)

Step4: Calculate the \(x\) - coordinate of the vertex

Using \(x=-\frac{b}{2a}\), we substitute \(a=-3\) and \(b = 6\):
\(x=-\frac{6}{2\times(-3)} = 1\)

Step5: Calculate the \(y\) - coordinate of the vertex

Substitute \(x = 1\) into \(f(x)=-3x^{2}+6x + 1\):
\(f(1)=-3\times1^{2}+6\times1+1=-3 + 6+1=4\)
The vertex is \((1,4)\)

Problem 7: \(f(x)=3(x + 2)(x + 6)\)

Here \(p=-2\), \(q=-6\)

Step6: Calculate the \(x\) - coordinate of the vertex

Using \(x=\frac{p + q}{2}\), we substitute \(p=-2\) and \(q=-6\):
\(x=\frac{-2+( - 6)}{2}=\frac{-8}{2}=-4\)

Step7: Calculate the \(y\) - coordinate of the vertex

Substitute \(x=-4\) into \(f(x)=3(x + 2)(x + 6)\):
\(f(-4)=3\times(-4 + 2)\times(-4+6)=3\times(-2)\times2=-12\)
The vertex is \((-4,-12)\)

Problem 8: \(f(x)=0.5(x - 2)(x + 6)\)

Here \(p = 2\), \(q=-6\)

Step8: Calculate the \(x\) - coordinate of the vertex

Using \(x=\frac{p + q}{2}\), we substitute \(p = 2\) and \(q=-6\):
\(x=\frac{2+( - 6)}{2}=\frac{-4}{2}=-2\)

Step9: Calculate the \(y\) - coordinate of the vertex

Substitute \(x=-2\) into \(f(x)=0.5(x - 2)(x + 6)\):
\(f(-2)=0.5\times(-2 - 2)\times(-2+6)=0.5\times(-4)\times4=-8\)
The vertex is \((-2,-8)\)

Problem 9: \(f(x)=-x(x + 6)\)

Here \(p = 0\), \(q=-6\)

Step10: Calculate the \(x\) - coordinate of the vertex

Using \(x=\frac{p + q}{2}\), we substitute \(p = 0\) and \(q=-6\):
\(x=\frac{0+( - 6)}{2}=-3\)

Step11: Calculate the \(y\) - coordinate of the vertex

Substitute \(x=-3\) into \(f(x)=-x(x + 6)\):
\(f(-3)=-(-3)\times(-3 + 6)=3\times3 = 9\)
The vertex is \((-3,9)\)

Answer:

Problem 5: Vertex: \((-3,-4)\)
Problem 6: Vertex: \((1,4)\)
Problem 7: Vertex: \((-4,-12)\)
Problem 8: Vertex: \((-2,-8)\)
Problem 9: Vertex: \((-3,9)\)