Finding the Rule of a Quadratic Function

||f(x)=ax^2||

1. Replace |x| and |f(x)| in the equation with the coordinates of a point.

2. Isolate parameter |a.|

3. Give the rule.

Find the rule of the quadratic function that passes through point |(-3,40.5).|

1. Replace |x| and |f(x)| in the equation with the coordinates of a point
   ||\begin{align}f(x)&=ax^2\\\color{#3a9a38}{40.5}&=a(\color{#3a9a38}{-3})^2\end{align}||

2. Isolate parameter |a|
   ||\begin{align}40.5&=a(-3)^2\\ 40.5&=9a\\\dfrac{40.5}{\color{#ec0000}9}&=\dfrac{9a}{\color{#ec0000}9}\\ 4.5&=a\end{align}||

3. Give the rule
   The function rule is |f(x)=4.5x^2.|

Find the rule of the quadratic function with a point at |(1.5,-11.25).|

1. Replace |x| and |f(x)| in the equation with the coordinates of a point
   ||\begin{align}f(x)&=ax^2\\\color{#3a9a38}{-11.25}&=a(\color{#3a9a38}{1.5})^2\end{align}||

2. Isolate parameter |a|
   ||\begin{align}-11.25&=a(1.5)^2\\ -11.25&=2.25a\\\dfrac{-11.25}{\color{#ec0000}{2.25}}&=\dfrac{2.25a}{\color{#ec0000}{2.25}}\\ -5&=a\end{align}||

3. Give the rule
   The function rule is |f(x)=-5x^2.|

||f(x)=a(x-h)^2+k|| where
|(h,k)|: coordinates of the vertex

1. Replace |h| and |k| in the equation with the coordinates of the vertex.

2. Replace |x| and |f(x)| in the equation with the coordinates of a point other than the vertex.

3. Isolate parameter |a.|

4. Give the rule.

Find the rule of the quadratic function with a vertex at coordinates |(4,6)| that passes through point |(2,-2).|

1. Replace |h| and |k| in the equation with the vertex’s coordinates
   ||\begin{align}f(x)&=a(x-h)^2+k\\ f(x)&=a(x-\color{#fa7921}4)^2+\color{#3b87cd}6\end{align}||

2. Replace |x| and |f(x)| in the equation with the coordinates of a point other than the vertex
   ||\begin{align}f(x)&=a(x-4)^2+6\\ \color{#3a9a38}{-2}&=a(\color{#3a9a38}2-4)^2+6\end{align}||

3. Isolate parameter |a|
   ||\begin{align}-2&=a(2-4)^2+6\\ -2&=a(-2)^2+6\\ -2&=4a+6\\ -8&=4a\\ -2&=a\end{align}||

4. Give the rule
   The function rule is |f(x)=-2(x-4)^2+6.|

Find the rule of the quadratic function with a vertex at the coordinates |(-1,2)| that passes through the point |(3,26).|

1. Replace |h| and |k| in the equation with the vertex’s coordinates
   ||\begin{align}f(x)&=a(x-h)^2+k\\ f(x)&=a(x-\color{#fa7921}{-1})^2+\color{#3b87cd}2\end{align}||

2. Replace |x| and |f(x)| in the equation with the coordinates of a point other than the vertex
   ||\begin{align}f(x)&=a(x+1)^2+2\\ \color{#3a9a38}{26}&=a(\color{#3a9a38}3+1)^2+2\end{align}||

3. Isolate parameter |a|
   ||\begin{align}26&=a(3+1)^2+2\\ 26&=a(4)^2+2\\ 26&=16a+2\\ 24&=16a\\ 1.5&=a\end{align}||

4. Give the rule
   The function rule is |f(x)=1.5(x+1)^2+2.|

||f(x)=a(x-x_1)(x-x_2)|| where
|x_1| and |x_2|: zeros (|x|-intercepts)

1. Replace |x_1| and |x_2| in the equation with the zeros.

2. Replace |x| and |f(x)| in the equation with the coordinates of a point other than the zeros.

3. Isolate parameter |a.|

4. Give the rule.

The order in which |x_1| and |x_2| are placed in the rule does not matter.

Find the rule of the quadratic function with zeros that are |-3| and |8|, that passes through the point |(5,-24).|

1. Replace |x_1| and |x_2| in the equation with the zeros
   ||\begin{align}f(x)&=a(x-x_1)(x-x_2)\\f(x)&=a(x-\color{#ff55c3}{-3})(x-\color{#ff55c3}8)\end{align}||

2. Replace |x| and |f(x)| in the equation with the coordinates of a point other than the zeros
   ||\begin{align}f(x)&=a(x+3)(x-8)\\\color{#3a9a38}{-24}&=a(\color{#3a9a38}5+3)(\color{#3a9a38}5-8)\end{align}||

3. Isolate parameter |a|
   ||\begin{align}-24&=a(5+3)(5-8)\\-24&=a(8)(-3)\\ -24&=-24a\\1&=a\end{align}||

4. Give the rule
   The function rule is |f(x)=(x+3)(x-8).|

Find the rule in general form of the quadratic function intersecting points |(-2,0),| |(7,18)| and |(3,0).|

It is not possible to directly determine the rule in the general form with the zeros, so we first find the rule in the factored form and then transform it.

1. Replace |x_1| and |x_2| in the equation with the zeros
   Use points |(-2,0)| and |(3,0)| to determine the zeros of the function, which are are |-2| and |3.|
   ||\begin{align}f(x)&=a(x-x_1)(x-x_2)\\f(x)&=a(x-\color{#ff55c3}{-2 })(x-\color{#ff55c3}3)\end{align}||

2. Replace |x| and |f(x)| in the equation with the coordinates of a point other than the zeros
   ||\begin{align}f(x)&=a(x+2)(x-3)\\\color{#3a9a38}{18}&=a(\color{#3a9a38}7+2)(\color{#3a9a38}7-3 )\end{align}||

3. Isolate parameter |a|
   ||\begin{align}18&=a(7+2)(7-3)\\18&=a(9)(4)\\18&=36a\\0.5&=a\end{align} ||

4. Give the rule
   The function rule in factored form is |f(x)=0.5(x+2)(x-3).|
   Expand the expression to transform it into general form.
   ||\begin{align}f(x)&=0.5(x+2)(x-3)\\&= 0.5(x^2-3x+2x-6)\\&= 0.5(x^2-x-6)\\&= 0.5x^2-0.5x-3 \end{align}||
   The rule of the function in general form is |f(x)=0.5x^2-0.5x-3.|

1. Use the formula to calculate the value of |h|.

2. Check if the third point provided has an |x|-coordinate equal to |h.| If yes, the point is the vertex. If not, refer to case 2.

3. Replace |h| and |k| in the standard equation with the vertex coordinates.

4. Replace |x| and |f(x)| in the equation with the coordinates of a point different from the vertex.

5. Isolate |a.|

6. Give the rule.

Find the equation of the quadratic function represented in the table of values below.

1. Calculate the value of |h|
   Points |(-4,4)| and |(2,4)| have the same |y|-coordinate, so we can calculate |h| from their |x|-coordinate.
   ||\begin{align}\color{#fa7921}h&=\dfrac{x_A+x_B}{2}\\ &=\dfrac{-4+2}{2}\\&=\dfrac{-2}{2}\\ &=\color{#fa7921}{-1}\end{align}||

2. Check if another point has an |x|-coordinate equal to |h|
   We notice that |-1| is the |x|-coordinate of one of the points in the table of values. We determine that |(-1,5)| is the vertex of the parabola, so |\color{#3b87cd}k=\color{#3b87cd}{-5}.|

3. Replace |h| and |k| in the equation with the coordinates of the vertex
   ||\begin{align}f(x)&=a(x-h)^2+k\\ f(x)&=a(x-\color{#fa7921}{-1})^2+\color{#3b87cd}{-5}\end{align}||

4. Replace |x| and |f(x)| in the equation with the coordinates of a point other than the vertex
   We can use the point |(-4,4).|||\begin{align}f(x)&=a(x+1)^2-5 \\ \color{#3a9a38}{4}&=a(\color{#3a9a38}{-4}+1)^2-5\end{align}||

5. Isolate |a|
   ||\begin{align}4&=a(-3)^2-5\\4&=9a-5\\9&=9a\\1&=a\end{align}||

6. Give the rule
   The rule of the function is |f(x)=(x+1)^2-5.|

1. Calculate the value of |h| using the formula.

2. Replace |h| in the equation with the value found in the previous step.

3. Create a system of equations by replacing |x| and |f(x)| with the coordinates of 2 points. The 2 points must not be the ones that have the same |y|-value.

4. Solve the system of equations to find the value of |a| and |k.|

5. Give the rule.

Find the equation of the following quadratic function.

1. Calculate the value of |h|
   ||\begin{align}\color{#fa7921}h&=\dfrac{x_A+x_B}{2}\\ &=\dfrac{-4+12}{2}\\&=\dfrac{8}{2}\\ &=\color{#fa7921}4\end{align}||

2. Replace |h| in the equation
   ||\begin{align}f(x)&= a(x-h)^2+k\\ &=a(x-\color{#fa7921}{4})^2+k \end{align}||

3. Create a system of equations with 2 points

4. Solve the system of equations to determine the value of |a| and |k|
   Start by isolating |k| in the first equation to use the substitution method.
   ||\begin{align}10&=64a+k\\\color{#3b87cd}{10-64a}&=\color{#3b87cd}k\end{align}||
   Then, replace |k| in the second equation with the found expression.
   ||\begin{align}13.5&=36a+\color{#3b87cd}k\\13.5&=36a+ \color{#3b87cd}{10-64a}\\13.5&=-28a+10\\3.5&=-28a\\\color{#3a9a38}{-0.125}&=\color{#3a9a38}a\end{align}||
   Last, determine the value of |k.| Use the equation where |k| is isolated.
   ||\begin{align}k&=10-64\color{#3a9a38}{a}\\&=10-64(\color{#3a9a38}{-0.125})\\&=10--8\\&=18\end{align}||

5. Give the rule
   The function rule is |f(x)=-0.125(x-4)^2+18.|