Question

find the vertex and axis of symmetry of each function. you are not required to graph here.

1. $f(x)=x^{2}+10x + 24$
2. $f(x)=x^{2}-4x - 21$
3. $f(x)=2x^{2}+16x + 30$
4. $f(x)=3x^{2}+15x + 30$
5. jessica is building a model rocket for her physics class. after studying the flight path of her rocket, she has concluded that she wants her rocket to achieve a maximum height of 50 ft. the equation for her rocket is $-3x^{2}+6x + 48$. will jessicas rocket clear 50 ft? (hint find the vertex of the equation to find the maximum height of the rocket)

optional: complete problems 5 - 17 from your textbook on separate lined paper.
when you are finished, check the solutions at the back of the textbook

Explanation:

Step1: Recall the formula for the vertex and axis of symmetry of a quadratic function

For a quadratic function \(y = ax^{2}+bx + c\), the axis of symmetry is given by the formula \(x=-\frac{b}{2a}\). The \(x\)-coordinate of the vertex is \(x =-\frac{b}{2a}\), and the \(y\)-coordinate of the vertex is \(y=f(-\frac{b}{2a})\).

Step2: Find the vertex and axis of symmetry for \(f(x)=x^{2}+10x + 24\)

For \(a = 1\), \(b=10\), \(c = 24\)

• Axis of symmetry:

Using the formula \(x=-\frac{b}{2a}\), substitute \(a = 1\) and \(b = 10\)
\(x=-\frac{10}{2\times1}=- 5\)

• Vertex:

First, find the \(x\)-coordinate \(x=-5\). Then find the \(y\)-coordinate by substituting \(x=-5\) into \(f(x)\)
\(f(-5)=(-5)^{2}+10\times(-5)+24=25-50 + 24=-1\)
The vertex is \((-5,-1)\)

Step3: Find the vertex and axis of symmetry for \(f(x)=x^{2}-4x-21\)

For \(a = 1\), \(b=-4\), \(c=-21\)

• Axis of symmetry:

Using the formula \(x =-\frac{b}{2a}\), substitute \(a = 1\) and \(b=-4\)
\(x=-\frac{-4}{2\times1}=2\)

• Vertex:

First, find the \(x\)-coordinate \(x = 2\). Then find the \(y\)-coordinate by substituting \(x = 2\) into \(f(x)\)
\(f(2)=2^{2}-4\times2-21=4-8-21=-25\)
The vertex is \((2,-25)\)

Step4: Find the vertex and axis of symmetry for \(f(x)=2x^{2}+16x + 30\)

For \(a = 2\), \(b = 16\), \(c = 30\)

• Axis of symmetry:

Using the formula \(x=-\frac{b}{2a}\), substitute \(a = 2\) and \(b = 16\)
\(x=-\frac{16}{2\times2}=-4\)

• Vertex:

First, find the \(x\)-coordinate \(x=-4\). Then find the \(y\)-coordinate by substituting \(x=-4\) into \(f(x)\)
\(f(-4)=2\times(-4)^{2}+16\times(-4)+30=2\times16-64 + 30=32-64+30=-2\)
The vertex is \((-4,-2)\)

Step5: Find the vertex and axis of symmetry for \(f(x)=3x^{2}+15x + 30\)

For \(a = 3\), \(b = 15\), \(c = 30\)

• Axis of symmetry:

Using the formula \(x=-\frac{b}{2a}\), substitute \(a = 3\) and \(b = 15\)
\(x=-\frac{15}{2\times3}=-\frac{5}{2}=-2.5\)

• Vertex:

First, find the \(x\)-coordinate \(x=-2.5\). Then find the \(y\)-coordinate by substituting \(x=-2.5\) into \(f(x)\)
\(f(-2.5)=3\times(-2.5)^{2}+15\times(-2.5)+30=3\times6.25-37.5 + 30=18.75-37.5+30=11.25\)
The vertex is \((-2.5,11.25)\)

Step6: Determine if Jessica's rocket will clear 50 ft

For the function \(y=-3x^{2}+6x + 48\), where \(a=-3\), \(b = 6\), \(c = 48\)

• Axis of symmetry:

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

• Vertex:

First, find the \(x\)-coordinate \(x = 1\). Then find the \(y\)-coordinate by substituting \(x = 1\) into \(y=-3x^{2}+6x + 48\)
\(y=-3\times(1)^{2}+6\times1+48=-3 + 6+48=51\)

Answer:

1. Axis of symmetry: \(x=-5\), Vertex: \((-5,-1)\)
2. Axis of symmetry: \(x = 2\), Vertex: \((2,-25)\)
3. Axis of symmetry: \(x=-4\), Vertex: \((-4,-2)\)
4. Axis of symmetry: \(x=-2.5\), Vertex: \((-2.5,11.25)\)
5. Since the maximum height (the \(y\)-coordinate of the vertex) of the rocket \(y = 51\) ft and \(51>50\), Jessica's rocket will clear 50 ft.