Solve $(x-2)^2-3=1$ graphically. Then use algebraic strategies to solve the equation and verify that your graphical solutions are correct.

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05-11-2026
Mathematics

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Solve $(x-2)^2-3=1$ graphically. Then use algebraic strategies to solve the equation and verify that your graphical solutions are correct.

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-05T00:35:34-05:00

Define functions $f(x) = (x-2)^2 - 3$ and $g(x) = 1$ to represent the equation graphically.
Plot the parabola and the horizontal line to find that they intersect at $x = 0$ and $x = 4$.
Isolate $x$ algebraically by adding 3 and taking the square root to confirm the solutions are $x = 0$ and $x = 4$.
The final solutions are $\boxed{x=0, x=4}$.

Explanation

Problem Analysis and Setup

To solve the equation $(x-2)^2 - 3 = 1$, we examine both graphical and algebraic approaches. The objective is to identify the values of $x$ that satisfy this equality. We represent the equation as a system of two functions: $f(x) = (x-2)^2 - 3$ and $g(x) = 1$. The solution corresponds to the $x$-coordinates where the graph of the parabola $f(x)$ intersects the horizontal line $g(x)$.

Graphical Solution

To find the graphical solution, we analyze the function $f(x) = (x-2)^2 - 3$. This is a parabola that opens upward with its vertex at $(2, -3)$. By evaluating the function at various points, we can plot it:

At $x = 0$, $f(0) = (0-2)^2 - 3 = 4 - 3 = 1$.
At $x = 1$, $f(1) = (1-2)^2 - 3 = 1 - 3 = -2$.
At $x = 2$, $f(2) = (2-2)^2 - 3 = 0 - 3 = -3$.
At $x = 3$, $f(3) = (3-2)^2 - 3 = 1 - 3 = -2$.
At $x = 4$, $f(4) = (4-2)^2 - 3 = 4 - 3 = 1$.

The line $g(x) = 1$ is a horizontal line passing through $y = 1$. By observing the plotted points, the graph of the parabola intersects the line $g(x) = 1$ exactly at $x = 0$ and $x = 4$.

Algebraic Solution and Verification

To solve the equation $(x-2)^2 - 3 = 1$ algebraically, we use inverse operations to isolate $x$:

Add 3 to both sides: $(x-2)^2 = 1 + 3$, which simplifies to $(x-2)^2 = 4$.
Take the square root of both sides: $x - 2 = \pm\sqrt{4}$. This gives $x - 2 = 2$ or $x - 2 = -2$.
Solve each linear equation:

For $x - 2 = 2$, we add 2 to both sides, resulting in $x = 4$.
For $x - 2 = -2$, we add 2 to both sides, resulting in $x = 0$.

The algebraic solutions are $x = 0$ and $x = 4$. These match the values identified from the graphical analysis, confirming that our findings are correct.

Examples

Graphical and algebraic solving methods are fundamental in fields like physics and engineering. For instance, when designing the trajectory of a projectile, an engineer might model the height as a quadratic function of time. By setting the function equal to a specific height (like the height of a target), the engineer can solve for the times when the projectile will pass through that height. Mastering these methods allows for precise calculations of physical phenomena.