Question

describe how to graph the solution of
y ≤ -x² + 2x.

in your description, did you
□ graph the parabola opening down
with the vertex at (1, 1)?
□ factor the quadratic as y ≤-1(x)(x – 2)?
□ find the roots of the parabola to be 0 and 2?
□ graph the parabola with a solid boundary
line?
□ test a point that is not on the boundary?
□ shade inside the parabola?

Explanation:

Step1: Analyze the quadratic function

The inequality is \( y \leq -x^{2}+2x \). First, consider the quadratic function \( y = -x^{2}+2x \). The coefficient of \( x^{2} \) is -1, which is negative, so the parabola opens downwards. To find the vertex, we can complete the square or use the formula \( x = -\frac{b}{2a} \). For \( y=-x^{2}+2x \), \( a=-1 \), \( b = 2 \), so \( x=-\frac{2}{2\times(-1)} = 1 \). Substitute \( x = 1 \) into the equation: \( y=-(1)^{2}+2\times1=-1 + 2=1 \), so the vertex is at \( (1,1) \).

Step2: Factor the quadratic

Factor \( -x^{2}+2x \): \( -x^{2}+2x=-x(x - 2)=-1\times x\times(x - 2) \), so \( y\leq-1(x)(x - 2) \) is a correct factoring.

Step3: Find the roots

To find the roots, set \( y = 0 \): \( 0=-x^{2}+2x \), \( x(-x + 2)=0 \), so \( x = 0 \) or \( -x+2=0\Rightarrow x = 2 \). So the roots are 0 and 2.

Step4: Determine the boundary line

The inequality is \( y\leq -x^{2}+2x \), the "less than or equal to" sign means we use a solid boundary line because the points on the parabola are included in the solution.

Step5: Test a point

To determine which region to shade, we test a point not on the boundary, for example, the origin \( (0,0) \). Substitute into the inequality: \( 0\leq-(0)^{2}+2\times0\Rightarrow0\leq0 \), which is true. But we can also test other points. The process of testing a point not on the boundary is part of graphing the solution of an inequality.

Step6: Shade the region

Since the parabola opens downwards and the test point (e.g., origin) satisfies the inequality, we shade the region inside (below, since the parabola opens down) the parabola.

Now, for the check - list:

• Graph the parabola opening down with the vertex at \( (1,1) \): Yes, as we found the vertex at \( (1,1) \) and the parabola opens down.
• Factor the quadratic as \( y\leq-1(x)(x - 2) \): Yes, from the factoring step.
• Find the roots of the parabola to be 0 and 2: Yes, from the root - finding step.
• Graph the parabola with a solid boundary line: Yes, because the inequality is \( \leq \), so the boundary is solid.
• Test a point that is not on the boundary: Yes, this is a standard step in graphing inequalities.
• Shade inside the parabola: Yes, because the parabola opens down and the test point inside (relative to the parabola's opening) satisfies the inequality.

To graph the solution of \( y\leq -x^{2}+2x \):

1. First, graph the parabola \( y=-x^{2}+2x \):

• Plot the vertex \( (1,1) \).
• Plot the roots \( (0,0) \) and \( (2,0) \).
• Since the coefficient of \( x^{2} \) is negative, draw a parabola opening downwards with a solid line (because of \( \leq \)).

1. Test a point not on the parabola, e.g., \( (0,0) \): substitute into \( y\leq -x^{2}+2x \), we get \( 0\leq0 \), which is true.
2. Shade the region that includes the test point, which is the region inside (below) the parabola.

Answer:

To graph \( y\leq -x^{2}+2x \):

1. Recognize the parabola \( y = -x^{2}+2x \) opens down (vertex at \( (1,1) \), roots at \( x = 0,2 \)), draw it with a solid line.
2. Test a non - boundary point (e.g., \( (0,0) \)): \( 0\leq0 \) (true), so shade the region inside (below) the parabola.

For the check - list, all the statements are correct:

• graph the parabola opening down with the vertex at \( (1,1) \): Yes
• factor the quadratic as \( y\leq-1(x)(x - 2) \): Yes
• find the roots of the parabola to be 0 and 2: Yes
• graph the parabola with a solid boundary line: Yes
• test a point that is not on the boundary: Yes
• shade inside the parabola: Yes