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Finding the \( x \)-intercepts is all about locating where the curve hits the \( x \)-axis. At this point, the value of \( y \) will equal zero because the curve is not above or below the \( x \)-axis.To derive this in parametric equations, you set the \( y \)-equation to zero, allowing you to solve for \( t \). This particular exercise gives us the quadratic equation \( t^2 + t - 6 = 0 \). We factor it into \((t-2)(t+3) = 0\) to find \( t = 2 \) or \( t = -3 \).Using these \( t \)-values in the equation for \( x \), \( x=t^2+t \), helps us find the exact coordinates:

• For \( t = 2 \), \( x = 6 \).
• For \( t = -3 \), \( x = 6 \).

Both yield the point \((6, 0)\) on the curve, so all \( x \)-intercepts merge into \((6, 0)\). This reflects that the curve just touches the \( x \)-axis there, not crossing it at distinct spots.

The \( y \)-intercepts are points where the curve crosses the \( y \)-axis. Here, \( x \) is zero. To find these in parametric equations, you set the \( x \) equation equal to zero and solve for \( t \).In the problem, solving the equation \( t^2 + t = 0 \) resulting from \( x = 0 \) gives \( t = 0 \) or \( t = -1 \).Next, substituting back into the \( y \) equation \( y=t^2+t-6 \):

• For \( t = 0 \), \( y = -6 \).
• For \( t = -1 \), \( y = -6 \).

This provides the same \( y \)-intercept, \((0, -6)\), for both \( t \)-values. Like the \( x \)-intercept, all possible \( y \)-intercepts collapse into one point.

Quadratic equations are polynomial equations of degree two and they appear as the standard form \( ax^2 + bx + c = 0 \). In this exercise, we see them in two different contexts due to parametric equations.For \( y=t^2+t-6 \), we set \( y=0 \) to solve for \( x \)-intercepts and get \( t^2 + t - 6 = 0 \). Factoring it gives \( (t-2)(t+3)=0 \), helping us find \( t \) values.Similarly, setting \( x=t^2 + t = 0 \), a simpler quadratic form, helps us find \( y \)-intercepts.What makes quadratics interesting is how every solution portrays a curve line shape or parabola. The solutions reflect critical points like the apex, intercepts, and crossing points. Understanding these allows us to pinpoint precisely where curves intersect the \( x \)- and \( y \)-axes in parametric equations.For this problem:

• Each solution determines how these parametric equations translate into the Cartesian plane.
• Solving the quadratic gives us specific parameter values that highlight various positions of the curve.

Grasping quadratic behavior is key to analyzing intersect points comprehensively.