Questions: Lesson 3.1 Homework Question 22 of 35 Question 22 For the equation (y=x^2-25), (a) Determine whether the parabola opens upward or downward. (b) Find the vertex. (c) Find the (y)-intercept. (d) Sketch the graph.

Solution Steps

The given equation is \( y = x^2 - 25 \). This is a quadratic equation in the form \( y = ax^2 + bx + c \), where \( a = 1 \), \( b = 0 \), and \( c = -25 \).

Since \( a = 1 \) is positive, the parabola opens upward.

The vertex of a parabola in the form \( y = ax^2 + bx + c \) is given by the formula: \[ x = -\frac{b}{2a} \] Substituting the values of \( a \) and \( b \): \[ x = -\frac{0}{2 \times 1} = 0 \] Substitute \( x = 0 \) back into the equation to find \( y \): \[ y = (0)^2 - 25 = -25 \] Thus, the vertex is at \( (0, -25) \).

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the equation: \[ y = (0)^2 - 25 = -25 \] Thus, the y-intercept is \( (0, -25) \).

Final Answer