Finding the Equation of a Parabola

 

1) \( \quad y = ax^2 + bx + c \)

 

 

If any three points on the parabola are given, we can substitute these coordinates into the general form of the quadratic equation to determine the coefficients $a, b,$ and $c$, thereby finding the equation of the parabola.

 

 

 

 

Example:

 

 

Let us find the equation of the parabola shown in the figure.

 

 

 

 

 

By substituting the coordinates of the given points on the parabola into the general form

\[
y = ax^2 + bx + c
\]

we obtain the following system of linear equations:

For \( A(-1 , 5)  \):  \( a – b + c = 5 \),

For \(B(2 , 5) \):  \( 4a + 2b + c = 5 \), and

For \( C(1 , 3) \):  \( a + b + c = 3 \).

Solving this system of equations simultaneously yields:

\[ a = 1,  \quad   b = -1 \quad  \text{and }  c = 3 \]

Therefore, the equation of the parabola is:

\[ y = ax^2 + bx + c \Rightarrow y = x^2 – x + 3 \]

 

2)   If the x-intercepts of the parabola and any additional point on it are given, its equation can be found using the factored form:

\[ y = a(x – x_1)(x – x_2) \]

 

Example:

 

Let us determine the equation of the parabola shown above. Since the x-intercepts are located at

\[ x_1 = -1 \quad  \text{and }  x_2 = 2,   \]

we substitute them into the factored form expression:

\[ y = a(x + 1)(x – 2) \]

To find the value of the leading coefficient \( a \), we substitute the given point \( (-2, \frac{5}{3}) \) into this equation:

For \( x = -2 \): \( y = \frac{5}{3} = a(-2 + 1)(-2 – 2) \)

\[ \Rightarrow a = \frac{5}{12} \quad  \text{and the equation is }  y = \frac{5}{12} (x + 1)(x – 2). \]

 

3) If the vertex of the parabola and any other point on it are given, its equation can be found using the vertex form:

\[ y = a(x – r)^2 + k \]

 

Example:

 

 

 

Let us find the equation of the parabola with vertex T shown in the figure.

 

 

 

 

 

Since the coordinates of the vertex are:

\[ r = -1 \quad  \text{and }  \quad    k = 4, \]

substituting these into the vertex form \( y = a(x – r)^2 + k \) yields:

\[ y = a(x + 1)^2 + 4 \]

To find the value of \( a \), we substitute the y-intercept \( (0, 3) \) into this equation:

\[ \text{For } x = 0: \quad   y = 3 = a(0 + 1)^2 + 4 \]

\[ \Rightarrow a = -1  \quad  \text{and the equation is }   y = -(x + 1)^2 + 4. \]

 

QUESTION 13

 

In the figure above, the parabola \( y = f(x) \) with vertex T passes through the origin. What is the value of \( f(5) \)?

\[
\text{A)} -6 \quad
\text{B) } -5 \quad
\text{C) } -4 \quad
\text{D) } -3 \quad
\text{E) } -2
\]

 

Solution:

 

Since the parabola intersects the x-axis at

\[ x_1 = 0 \quad  \text{and } \quad    x_2 = 4, \]

we can express it in factored form as \( y = a(x – x_1)(x – x_2) \Rightarrow y = a x (x – 4) \). The x-coordinate of the vertex is given by the axis of symmetry:

\[
r = \frac{x_1 + x_2}{2} \Rightarrow r = \frac{0 + 4}{2} = 2.
\]

Thus, the vertex is \( T(2, 4) \). Substituting this point into our equation:

\[ \text{For } x = 2: \quad   y = 4 = a \cdot 2 \cdot (2 – 4) \]

\[   \Rightarrow a = -1 \quad  \text{which gives }   f(x) = -x(x – 4).   \]

Evaluating the function at \( x = 5 \):

\[ f(5) = -5(5 – 4) = -5 \]

\(\textbf{Correct Answer: B} \)

 

QUESTION 14

 

The figure above shows a parabola with vertex \( T(-2, -1) \). What is the area of triangle ATB in square units?

\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]

 

Solution:

 

Given the vertex coordinates \( r = -2 \) and \( k = -1 \), the equation in vertex form is:

\[
y = a(x – r)^2 + k \Rightarrow y = a(x + 2)^2 – 1
\]

Substituting the y-intercept \( (0, 3) \) into the equation to solve for \( a \):

\[
\text{For } x = 0: \quad y = 3 = a(0 + 2)^2 – 1
\]

\[
\Rightarrow a = 1 \text{ and } y = (x + 2)^2 – 1
\]

Next, let us find the x-intercepts of the parabola by setting \( y = 0 \):

\[
y = (x + 2)^2 – 1 = 0 \Rightarrow x^2 + 4x + 3 = 0
\]

\[
\Rightarrow x = -3 \text{ or } x = -1
\]

This gives the points \( A(-3, 0) \) and \( B(-1, 0) \). The base of triangle ATB lies along the x-axis, and its length is:

\[
|AB| = |-1 – (-3)| = 2
\]

The height of the triangle is equal to the absolute value of the y-coordinate of the vertex \( T \):

\[
h = |k| = |-1| = 1
\]

Therefore, the area of triangle ATB is:

\[
\text{Area}(ATB) = \frac{ |AB| \cdot |k| }{2} = \frac{2 \cdot |-1|}{2} = 1 \text{ square unit.}
\]

 

\(\textbf{Correct Answer: A} \)

 

 

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