How to Find the Inverse of a Quadratic Function

Graph f(x) = x². Your calculator shows a symmetric curve. Ask for the inverse. It shows two curves, one above y=x and one below, and it calls them both f^-1(x). That is the problem. The inverse of a quadratic function, taken on all real numbers, fails the horizontal line test. Every horizontal line above the vertex hits the curve twice. A function that maps one x to two y is not a function. The correct inverse of a quadratic exists only after you restrict the domain to a single side of the vertex.

A function must be one-to-one to have an inverse function. A one-to-one function passes the horizontal line test: no horizontal line touches the graph more than once. Every quadratic f(x)=ax²+bx+c, with a≠0, is symmetric. The left side and the right side produce the same output y for two different x inputs. This fails the horizontal line test, so no quadratic has an inverse function over its entire domain of all real numbers. The two branches exist as relations, but neither is a function alone. Cut the curve into two halves. Each half is one-to-one, and each half has its own inverse function. The cut always goes at the vertex.

The vertex of a quadratic is the turning point. For f(x)=ax²+bx+c, the x-coordinate of the vertex is x=-b/(2a). To restrict the domain for a quadratic inverse, choose either the left side of the vertex (x ≤ -b/(2a)) or the right side (x ≥ -b/(2a)). Both sides are one-to-one. Which one you pick determines the range of the inverse. The standard choice for f(x)=x² is the right side, x ≥ 0, giving the inverse f^-1(x)=√x. For f(x)=x²-4, the vertex is at x=0. The right branch x ≥ 0 is one-to-one, and the left branch x ≤ 0 is one-to-one. The domain restriction is not optional; it is the only way to produce a valid function inverse.

The domain of the restricted function becomes the range of its inverse. The range of the restricted function becomes the domain of its inverse. For f(x)=x² with domain [0,∞), the range is [0,∞). The inverse √x has domain [0,∞) and range [0,∞). This swap is the inverse function domain range interpretation in action. You cannot skip it without creating an inverse that is not a function.

To find the inverse of a quadratic that is not a simple x², rewrite it in vertex form. The completing the square inverse method is the standard algebra procedure. Write f(x)=ax²+bx+c as f(x)=a(x-h)²+k, where (h,k) is the vertex. Then swap x and y, and solve for y. The steps are direct.

Take f(x)=x²+6x+5. Complete the square: f(x)=(x+3)²-4. The vertex is at (-3,-4). Swap x and y: x=(y+3)²-4. Solve for y: x+4=(y+3)². Then y+3=±√(x+4). The inverse is y=-3±√(x+4). With the domain restricted to x≥-3 (the right side), the inverse is f^-1(x)=-3+√(x+4). With the domain restricted to x≤-3, the inverse is f^-1(x)=-3-√(x+4). The completing the square inverse method works every time because it isolates the squared term.

When Completing the Square Is Required

If the quadratic has a linear term b≠0, you cannot simply take the square root. The inverse of x² is ±√x, but the inverse of x²+6x+5 requires the full completing the square inverse procedure. Any quadratic that is not a perfect square or a simple x² needs this method. OpenStax Precalculus 2e covers this exact procedure in Section 3.7, Inverse Functions. The 2021 edition from OpenStax at Rice University uses the same algebra.

The quadratic formula inverse method is faster for some students. Start with the general form f(x)=ax²+bx+c. Swap x and y: x=ay²+by+c. Rearrange to ay²+by+(c-x)=0. This is a quadratic in y. Apply the quadratic formula: y=[-b±√(b²-4a(c-x))]/(2a). The expression under the square root is the discriminant. This gives the inverse directly, without completing the square.

For f(x)=2x²+4x-3, swap to get x=2y²+4y-3. Rearrange: 2y²+4y+(-3-x)=0. Use the quadratic formula: y=[-4±√(16-4(2)(-3-x))]/(4). Simplify the discriminant: 16-8(-3-x)=16+24+8x=40+8x. So y=[-4±√(40+8x)]/4 = -1±(√(40+8x))/4. Factor: y=-1±(√(8(5+x)))/4 = -1±(2√(2(5+x)))/4 = -1±(√(2(5+x)))/2. This is the inverse relation. The vertex is at x=-b/(2a)=-4/(4)=-1. Restrict the domain to x≥-1 or x≤-1. Choose the + branch for x≥-1: f^-1(x)=-1+(√(2(5+x)))/2. The quadratic formula inverse method works for any quadratic, but it requires careful simplification.

The ± in the inverse is not a choice of convenience. It is determined by the domain restriction you applied to the original function. The vertex is the boundary. If you restricted the domain to x ≥ vertex (the right branch), choose the + branch when a>0. If you restricted to x ≤ vertex (the left branch), choose the - branch when a>0. The sign flips when a is negative.

For f(x)=x²-4, vertex at x=0. Right branch x≥0: inverse is +√(x+4). Left branch x≤0: inverse is -√(x+4). For f(x)=-x²+4, vertex at x=0. Right branch x≥0: inverse is -√(4-x). Left branch x≤0: inverse is +√(4-x). Test this by plugging a point from the restricted branch into the inverse. If f(2)=0 for a quadratic, then f^-1(0) must return 2, not -2. The branch that gives 2 is the correct one.

These examples show the full process, including the graph of the restricted branch. Each ends with verification by composition.

Example 1: f(x) = x² - 4x + 7

Vertex: x = -b/(2a) = 4/(2) = 2. Restrict domain to x ≥ 2. Complete the square: f(x) = (x-2)² + 3. Swap x and y: x = (y-2)² + 3. Solve: (y-2)² = x - 3. y - 2 = ±√(x-3). y = 2 ± √(x-3). For the right branch (x≥2, y≥3), choose +. The inverse is f^-1(x) = 2 + √(x-3). Domain of inverse: x ≥ 3. Range: y ≥ 2. Verify: f(f^-1(x)) = (2+√(x-3)-2)² + 3 = (√(x-3))² + 3 = x-3+3 = x. Verified.

Example 2: f(x) = 3x² + 6x - 2

Vertex: x = -6/(6) = -1. Restrict domain to x ≥ -1. Use the quadratic formula method. Swap: x = 3y² + 6y - 2. Rearrange: 3y² + 6y + (-2 - x) = 0. y = [-6 ± √(36 - 12(-2-x))]/6 = [-6 ± √(36+24+12x)]/6 = [-6 ± √(60+12x)]/6 = [-6 ± √(12(5+x))]/6 = [-6 ± 2√(3(5+x))]/6 = -1 ± (√(3(5+x)))/3. For x ≥ -1, the range of f is y ≥ f(-1) = 3(1)-6-2 = -5. Choose the + branch: f^-1(x) = -1 + (√(3(5+x)))/3. Domain: x ≥ -5. Verify: f(f^-1(x)) = 3(-1 + √(3(5+x))/3)² + 6(-1 + √(3(5+x))/3) - 2. This simplifies to x. The composition check is the only reliable verification; a calculator cannot do it for you.

Example 3: f(x) = -x² + 2x + 3

Vertex: x = -2/(-2) = 1. a is negative. Restrict to x ≤ 1 (left side). Complete the square: f(x) = -(x² - 2x) + 3 = -(x² - 2x + 1 - 1) + 3 = -((x-1)² - 1) + 3 = -(x-1)² + 1 + 3 = -(x-1)² + 4. Swap: x = -(y-1)² + 4. Solve: (y-1)² = 4 - x. y - 1 = ±√(4-x). y = 1 ± √(4-x). For the left branch x ≤ 1, the range of f is y ≤ 4. The inverse must map back to values ≤ 1. Choose -: f^-1(x) = 1 - √(4-x). Domain: x ≤ 4. Range: y ≤ 1. Verify: f(f^-1(x)) = -(1-√(4-x)-1)² + 4 = -(-√(4-x))² + 4 = -(4-x) + 4 = x. Verified.

Graph f(x)=x²-4x+7 for x ≥ 2. The curve opens upward, vertex at (2,3). Draw the line y=x. The inverse function f^-1(x)=2+√(x-3) appears as the reflection of this half across y=x. The inverse starts at (3,2) and curves upward to the right. If you graph the left branch x≤2, its reflection would be below y=x. The two inverse branches together form the full sideways curve, but only one is the function inverse of your restricted domain. OpenStax Precalculus 2e Section 5.3, Inverse Trigonometric Functions, uses the same principle of cutting the domain to define arcsin, arccos, and arctan. The cut of sin(x) to [-π/2, π/2] is mathematically identical to cutting a quadratic to one side of its vertex.