Question

plot the given parabola on the axes. plot the roots, the vertex and two other points.
y = -x² - 2x + 35
set the scales and drag the axes to change the graph.
xsel: 1 ysel: 1
done with scale

Explanation:

Step1: Find the roots (solve \( -x^2 - 2x + 35 = 0 \))

Multiply both sides by -1: \( x^2 + 2x - 35 = 0 \)
Factor: \( (x + 7)(x - 5) = 0 \)
So roots are \( x = -7 \) (when \( x + 7 = 0 \)) and \( x = 5 \) (when \( x - 5 = 0 \)).
Corresponding points: \( (-7, 0) \) and \( (5, 0) \).

Step2: Find the vertex (using \( x = -\frac{b}{2a} \) for \( y = ax^2 + bx + c \))

For \( y = -x^2 - 2x + 35 \), \( a = -1 \), \( b = -2 \).
\( x = -\frac{-2}{2(-1)} = -\frac{2}{2} = -1 \).
Substitute \( x = -1 \) into \( y \):
\( y = -(-1)^2 - 2(-1) + 35 = -1 + 2 + 35 = 36 \).
Vertex: \( (-1, 36) \).

Step3: Find two other points (e.g., \( x = 0 \) and \( x = 1 \))

• At \( x = 0 \): \( y = -0 - 0 + 35 = 35 \). Point: \( (0, 35) \).
• At \( x = 1 \): \( y = -1 - 2 + 35 = 32 \). Point: \( (1, 32) \).

To plot:

• Roots: \( (-7, 0) \), \( (5, 0) \)
• Vertex: \( (-1, 36) \)
• Other points: \( (0, 35) \), \( (1, 32) \) (or any two points, e.g., \( x = -2 \): \( y = -4 + 4 + 35 = 35 \) → \( (-2, 35) \))

Answer:

• Roots: \( (-7, 0) \), \( (5, 0) \)
• Vertex: \( (-1, 36) \)
• Two other points: \( (0, 35) \), \( (1, 32) \) (or similar valid points)

(Plot these on the coordinate axes with appropriate scales.)