Worked examples and practice exercises — using intercepts, the vertex, three points, and reflectors
Several methods can be used to find the equation of a parabola from its graph. The right method depends on the features visible on the graph — the vertex, the x-intercepts, or any three distinct points.
Below are four worked examples with detailed solutions, followed by three practice exercises.
| What you know | Form to use | How to solve |
|---|---|---|
| Two x-intercepts + one other point | \( y = a(x - r_1)(x - r_2) \) | Substitute the other point, solve for \( a \) |
| Vertex + one other point | \( y = a(x - h)^2 + k \) | Substitute the other point, solve for \( a \) |
| Three arbitrary points | \( y = ax^2 + bx + c \) | Solve a 3×3 linear system for \( a, b, c \) |
| Parabolic reflector (diameter \(D\), depth \(d\)) | \( y = \dfrac{x^2}{4p} \) (vertex at origin) | Use \( (D/2,\, d) \), solve for \( p \) |
Find the equation of the parabola whose graph is shown below.
The graph crosses the x-axis at \( x = -1 \) and \( x = 2 \), so the equation has the factored form:
\[ y = a(x + 1)(x - 2) \]Use the y-intercept \( (0, -2) \) to find \( a \):
\[ -2 = a(0 + 1)(0 - 2) \] \[ -2 = -2a \] \[ a = 1 \]Expanding:
\[ y = (x + 1)(x - 2) = x^2 - x - 2 \]Answer: \( y = x^2 - x - 2 \)
Find the equation of the parabola whose graph is shown below.
The vertex is \( (2, 3) \), so the vertex form is:
\[ y = a(x - 2)^2 + 3 \]Substitute the point \( (0, -1) \) to solve for \( a \):
\[ -1 = a(0 - 2)^2 + 3 \] \[ -1 = 4a + 3 \] \[ -4 = 4a \] \[ a = -1 \]Answer: \( y = -(x - 2)^2 + 3 \)
Check: the vertex is \( (2, 3) \) ✓, the y-intercept is \( -(0-2)^2 + 3 = -1 \) ✓, and the parabola opens downward (matching the graph).
Find the equation of the parabola whose graph is shown below.
Write the parabola with a vertical axis in standard form:
\[ y = ax^2 + bx + c \]Three points on the graph are \( (-1, 3) \), \( (0, -2) \), and \( (2, 6) \). Substituting:
\[ \begin{aligned} a - b + c &= 3 \\ c &= -2 \\ 4a + 2b + c &= 6 \end{aligned} \]Substituting \( c = -2 \) into the first and third equations:
\[ \begin{aligned} a - b - 2 &= 3 \;\Rightarrow\; a - b = 5 \\ 4a + 2b - 2 &= 6 \;\Rightarrow\; 2a + b = 4 \end{aligned} \]Adding the two equations gives \( 3a = 9 \), so \( a = 3 \). Then \( a - b = 5 \) gives \( b = -2 \).
Answer: \( y = 3x^2 - 2x - 2 \)
Check: \( f(-1) = 3 \), \( f(0) = -2 \), \( f(2) = 6 \) ✓
Find the equation of a parabolic reflector with diameter \( D = 2.3 \) m and depth \( d = 0.35 \) m, and determine the coordinates of its focus.
Place the vertex at the origin, so the equation has the form:
\[ y = \frac{1}{4p} x^2 \]A point on the rim of the reflector is at \( \left(\dfrac{D}{2}, d\right) = (1.15, 0.35) \). Substitute:
\[ 0.35 = \frac{(1.15)^2}{4p} \] \[ 4p = \frac{1.15^2}{0.35} = \frac{1.3225}{0.35} \approx 3.7786 \] \[ p \approx 0.9447 \text{ m} \]So \( \dfrac{1}{4p} \approx 0.2646 \), giving:
\[ y \approx 0.26 x^2 \]For an upward-opening parabola with vertex at the origin, the focus lies at \( (0, p) \):
\[ F(0,\; 0.94) \]Answer: \( y \approx 0.26 x^2 \) with focus at \( (0, 0.94) \).
Find the equation of the parabola in each graph. Expand View Answer to check your work.
The graph shows three labeled points: \( (0, -3) \), \( (2, -5) \), and \( (4, 1) \). Write the parabola in standard form:
\[ y = ax^2 + bx + c \]Substitute each point:
\[ \begin{aligned} a(0)^2 + b(0) + c &= -3 \\ a(2)^2 + b(2) + c &= -5 \\ a(4)^2 + b(4) + c &= 1 \end{aligned} \]Simplify:
\[ \begin{aligned} c &= -3 \\ 4a + 2b - 3 &= -5 \;\Rightarrow\; 2a + b = -1 \\ 16a + 4b - 3 &= 1 \;\Rightarrow\; 4a + b = 1 \end{aligned} \]Subtract the first simplified equation from the second:
\[ (4a + b) - (2a + b) = 1 - (-1) \;\Rightarrow\; 2a = 2 \;\Rightarrow\; a = 1 \]Substitute \( a = 1 \) into \( 4a + b = 1 \): \( 4 + b = 1 \Rightarrow b = -3 \).
Answer: \( y = x^2 - 3x - 3 \)
Check: \( f(0) = -3 \) ✓, \( f(2) = 4 - 6 - 3 = -5 \) ✓, \( f(4) = 16 - 12 - 3 = 1 \) ✓
Equation:
\[ y = -(x + 2)^2 - 1 = -x^2 - 4x - 5 \]Hint: the vertex is \( (-2, -1) \) and the parabola opens downward.
The graph crosses the x-axis at \( x = -3 \) and \( x = 2 \), so the equation has the factored form:
\[ y = a(x + 3)(x - 2) \]Use the y-intercept \( (0, -6) \) to find \( a \):
\[ -6 = a(0 + 3)(0 - 2) \] \[ -6 = a(3)(-2) \] \[ -6 = -6a \] \[ a = 1 \]Expanding:
\[ y = (x + 3)(x - 2) = x^2 + x - 6 \]Answer: \( y = x^2 + x - 6 \)
Check: \( f(-3) = 0 \) ✓, \( f(2) = 0 \) ✓, \( f(0) = -6 \) ✓
Identify which features are visible. With x-intercepts and one other point, use factored form \( y = a(x-r_1)(x-r_2) \). With the vertex and one other point, use vertex form \( y = a(x-h)^2 + k \). With only three arbitrary points, substitute into \( y = ax^2 + bx + c \) and solve a 3×3 system.
Vertex form \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex. Substitute the other point to find \( a \).
Place the vertex at the origin so the equation is \( y = x^2/(4p) \). A rim point has coordinates \( (D/2, d) \), where \( D \) is diameter and \( d \) is depth. Substituting that point and solving for \( p \) gives the equation and places the focus at \( (0, p) \).
Three non-collinear points determine a unique parabola with a vertical axis. If the three points are collinear, no such parabola exists.
Last updated: October 17, 2026