What is the formula for the vertex of a parabola with examples

The vertex of a parabola is the highest or lowest point on the curve, depending on its orientation. Its formula depends on the general form of the parabola’s equation.

1. Vertex Formula for a Parabola in Standard Form

For a quadratic equation in standard form:

y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+cThe x-coordinate of the vertex is given by:

x=−b2ax = -\frac{b}{2a}x=−2ab​To find the y-coordinate, substitute x=−b2ax = -\frac{b}{2a}x=−2ab​ back into the equation:

y=a(−b2a)2+b(−b2a)+cy = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + cy=a(−2ab​)2+b(−2ab​)+cThus, the vertex is at:

Vertex: (−b2a,y)\text{Vertex: } \left(-\frac{b}{2a}, y\right)Vertex: (−2ab​,y)

Example 1: Finding the Vertex of a Parabola in Standard Form

Find the vertex of the parabola:

y=2×2−4x+1y = 2x^2 – 4x + 1y=2x2−4x+1Step 1: Identify coefficients $a$, $b$, and $c$:

$$a=2,b=-4,c=1$$Step 2: Calculate the x-coordinate:

x=−b2a=−−42(2)=44=1x = -\frac{b}{2a} = -\frac{-4}{2(2)} = \frac{4}{4} = 1x=−2ab​=−2(2)−4​=44​=1Step 3: Substitute $x=1$ into the equation to find $y$:

y=2(1)2−4(1)+1=2−4+1=−1y = 2(1)^2 – 4(1) + 1 = 2 – 4 + 1 = -1y=2(1)2−4(1)+1=2−4+1=−1Result: The vertex is at $(1,-1)$.

2. Vertex Formula for a Parabola in Vertex Form

For a parabola in vertex form:

y=a(x−h)2+ky = a(x – h)^2 + ky=a(x−h)2+kThe vertex is directly given as:

$$\text{Vertex: }(h,k)$$

Example 2: Vertex in Vertex Form

Find the vertex of the parabola:

y=3(x−2)2+5y = 3(x – 2)^2 + 5y=3(x−2)2+5The equation is already in vertex form. Comparing it to y=a(x−h)2+ky = a(x – h)^2 + ky=a(x−h)2+k:

$$h=2,k=5$$Result: The vertex is at $(2,5)$.

3. Vertex Formula for a Parabola in General Form

For a parabola in general form:

Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0Ax2+Bxy+Cy2+Dx+Ey+F=0This case typically requires completing the square or advanced methods to rewrite the equation in standard or vertex form. Most commonly, the parabolas analyzed are already simplified to one of the forms above.

Key Observations

1. In standard form (y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c):
   • Use x=−b2ax = -\frac{b}{2a}x=−2ab​ to find the x-coordinate of the vertex.
2. In vertex form (y=a(x−h)2+ky = a(x – h)^2 + ky=a(x−h)2+k):
   • The vertex is directly at $(h,k)$.

Example 3: Graphical Interpretation

For the parabola y=−x2+4x−3y = -x^2 + 4x – 3y=−x2+4x−3:

1. $a=-1,b=4,c=-3$
2. Find x=−b2a=−42(−1)=2x = -\frac{b}{2a} = -\frac{4}{2(-1)} = 2x=−2ab​=−2(−1)4​=2.
3. Find $y$ by substituting $x=2$ into the equation: y=−22+4(2)−3=−4+8−3=1y = -2^2 + 4(2) – 3 = -4 + 8 – 3 = 1y=−22+4(2)−3=−4+8−3=1
4. Vertex: $(2,1)$.

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The vertex of a parabola is its highest or lowest point, depending on its orientation. The formula for the vertex depends on the equation of the parabola. Let’s break it down:

1. Parabola in Standard Form

The standard form of a parabola is:

y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c

Vertex Formula:

The x-coordinate of the vertex is:

x=−b2ax = -\frac{b}{2a}x=−2ab​The y-coordinate can be found by substituting x=−b2ax = -\frac{b}{2a}x=−2ab​ back into the equation:

y=a(−b2a)2+b(−b2a)+cy = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + cy=a(−2ab​)2+b(−2ab​)+cThus, the vertex is:

Vertex: (−b2a,y)\text{Vertex: } \left(-\frac{b}{2a}, y\right)Vertex: (−2ab​,y)

Example 1: Finding the Vertex of a Parabola in Standard Form

Find the vertex of the parabola:

y=2×2−4x+1y = 2x^2 – 4x + 1y=2x2−4x+1

1. Identify coefficients:

   $$a=2,b=-4,c=1$$

2. Compute the x-coordinate of the vertex:

   x=−b2a=−−42(2)=44=1x = -\frac{b}{2a} = -\frac{-4}{2(2)} = \frac{4}{4} = 1x=−2ab​=−2(2)−4​=44​=1

3. Compute the y-coordinate by substituting $x=1$ into the equation:

   y=2(1)2−4(1)+1=2−4+1=−1y = 2(1)^2 – 4(1) + 1 = 2 – 4 + 1 = -1y=2(1)2−4(1)+1=2−4+1=−1

Result: The vertex is at $(1,-1)$.

2. Parabola in Vertex Form

The vertex form of a parabola is:

y=a(x−h)2+ky = a(x – h)^2 + ky=a(x−h)2+k

Vertex Formula:

The vertex is directly given as:

$$\text{Vertex: }(h,k)$$

Example 2: Finding the Vertex in Vertex Form

Find the vertex of the parabola:

y=3(x−2)2+5y = 3(x – 2)^2 + 5y=3(x−2)2+5Here, the equation is already in vertex form. Comparing it to y=a(x−h)2+ky = a(x – h)^2 + ky=a(x−h)2+k:

$$h=2,k=5$$Result: The vertex is at $(2,5)$.

3. Parabola Opening Sideways (Horizontal Parabola)

For a parabola that opens sideways, the equation is in terms of $x$:

x=a(y−k)2+hx = a(y – k)^2 + hx=a(y−k)2+h

Vertex Formula:

The vertex is directly given as:

$$\text{Vertex: }(h,k)$$

Example 3: Horizontal Parabola

Find the vertex of the parabola:

x=2(y+3)2−1x = 2(y + 3)^2 – 1x=2(y+3)2−1Here, comparing it to x=a(y−k)2+hx = a(y – k)^2 + hx=a(y−k)2+h:

$$h=-1,k=-3$$Result: The vertex is at $(-1,-3)$.

Summary of Vertex Formulas

1. Standard Form (y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c):
   • Vertex: $(-b/2a,y)$, where $y$ is calculated by substituting $x=-b/2a$ into the equation.
2. Vertex Form (y=a(x−h)2+ky = a(x – h)^2 + ky=a(x−h)2+k):
   • Vertex: $(h,k)$.
3. Horizontal Parabola (x=a(y−k)2+hx = a(y – k)^2 + hx=a(y−k)2+h):
   • Vertex: $(h,k)$.

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