How is the point at $x = -1$ found on the curve $y = x^2 - 4$ ?

Q: How is the point at x = -1 found on the curve y = x^2 - 4?

To find the point on the curve at $x = -1$, substitute $x = -1$ into the curve's equation $y = x^2 - 4$. This gives $y = (-1)^2 - 4 = 1 - 4 = -3$. Therefore, the coordinates of the point are $(-1, -3)$.

To find the point on the curve at

x = -1

, substitute

x = -1

into the curve's equation

y = x^2 - 4

. This gives

y = (-1)^2 - 4 = 1 - 4 = -3

. Therefore, the coordinates of the point are

(-1, -3)

Conditions

The curve is defined by the equation $y = x^2 - 4$ .
The x-coordinate of the desired point is given as $-1$ .
The point must lie on the curve.

Reasoning, step by step

Identify the curve equation: $y = x^2 - 4$ .
Substitute the given x-value ( $x = -1$ ) into the equation.
Evaluate the square: $(-1)^2 = 1$ .
Perform the subtraction: $1 - 4 = -3$ .
Combine the x and y values to state the point coordinates: $P(-1, -3)$ .

Example

The video states: "For the first one, we know the x value, we don't know the y value, but we could find it by replacing x with negative one in that equation... $y = (-1)^2 - 4 = 1 - 4 = -3$ ."

Common misconceptions

Incorrectly evaluating $(-1)^2$ as $-1$ instead of $1$ .
Substituting $x = -1$ into the wrong equation or forgetting to subtract 4.
Assuming the y-value is 0 because it is an intercept (it is not an x-intercept).

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How is the point at x=−1x = -1x=−1 found on the curve y=x2−4y = x^2 - 4y=x2−4?