Graph the following quadratic functions, determine their domain and range: (a) y = -3x2 (b) y = -2x2 - 8 (c) y = x2 - 9 for $x ",

QUESTION 1 1 (a) y = -3x^2 Step 1: Create a table of values for x \-2, -1, 0, 1, 2\. |c|c| x & y = -3x^2 \\ -2 & -3(-2)^2 = -12 \\ -1 & -3(-1)^2 = -3 \\ 0 & -3(0)^2 = 0 \\ 1 & -3(1)^2 = -3 \\ 2 & -3(2)^2 = -12 \\ Step 2: Describe the graph, domain, and range. The graph is a parabola opening downwards with its vertex at (0,0). Domain: x R Range: y 0 1 (b) y = -2x^2 - 8 Step 1: Create a table of values for x \-2, -1, 0, 1, 2\. |c|c| x & y = -2x^2 - 8 \\ -2 & -2(-2)^2 - 8 = -16 \\ -1 & -2(-1)^2 - 8 = -10 \\ 0 & -2(0)^2 - 8 = -8 \\ 1 & -2(1)^2 - 8 = -10 \\ 2 & -2(2)^2 - 8 = -16 \\ Step 2: Describe the graph, domain, and range. The graph is a parabola opening downwards with its vertex at (0,-8). Domain: x R Range: y -8 1 (c) y = x^2 - 9 Step 1: Create a table of values for x \-2, -1, 0, 1, 2\. |c|c| x & y = x^2 - 9 \\ -2 & (-2)^2 - 9 = -5 \\ -1 & (-1)^2 - 9 = -8 \\ 0 & (0)^2 - 9 = -9 \\ 1 & (1)^2 - 9 = -8 \\ 2 & (2)^2 - 9 = -5 \\ Step 2: Describe the graph, domain, and range. The graph is a parabola opening upwards with its vertex at (0,-9). Domain: x R Range: y -9 QUESTION 2 Given: y = 4x^2 - 8 2 (a) Shape of the graph The graph is a parabola. Since the coefficient of x^2 is positive (4 > 0), it opens upwards. Shape: Parabola opening upwards 2 (b) x-intercept(s) Step 1: Set y = 0 to find the x-intercepts. 0 = 4x^2 - 8 Step 2: Solve for x. 4x^2 = 8 x^2 = (8)/(4) x^2 = 2 x = ± sqrt(2) x-intercepts: (sqrt(2), 0) and (-sqrt(2), 0) 2 (c) y-intercept Step 1: Set x = 0 to find the y-intercept. y = 4(0)^2 - 8 Step 2: Calculate y. y = -8 y-intercept: (0, -8) 2 (d) Turning point For a quadratic function of the form y = ax^2 + c, the turning point is (0, c). Turning point: (0, -8) 2 (e) Domain For any quadratic function, the domain is all real numbers. Domain: x R 2 (f) Range Since the parabola opens upwards and its turning point (minimum value) is at y = -8. Range: y -8 2 (g) Is the graph a function or a non-function? The graph of y = 4x^2 - 8 is a parabola that passes the vertical line test. It is a function. QUESTION 3 Calculate the roots of: 3 (a) y = 3x^2 Step 1: Set y = 0 to find the roots. 0 = 3x^2 Step 2: Solve for x. x^2 = (0)/(3) x^2 = 0 x = 0 Roots: x = 0 3 (b) y = -x^2 + 4 Step 1: Set y = 0 to find the roots. 0 = -x^2 + 4 Step 2: Solve for x. x^2 = 4 x = ± sqrt(4) x = ± 2 Roots: x = 2 and x = -2 3 (c) y = 2x^2 + 8 Step 1: Set y = 0 to find the roots. 0 = 2x^2 + 8 Step 2: Solve for x. 2x^2 = -8 x^2 = (-8)/(2) x^2 = -4 Since the square of a real number cannot be negative, there are no real roots. Roots: No real roots QUESTION 4 Given: f(x) = -3x^2 + 12 4 (a) Draw a sketch of f(x). Clearly indicate the coordinates of the intercepts on the axes as well as the turning point. Step 1: Determine the shape and turning point. The function is a parabola f(x) = ax^2 + c with a = -3 and c = 12. Since a = -3 < 0, the parabola opens downwards. The turning point is (0, c), so it is (0, 12). Step 2: Determine the y-intercept. Set x = 0: f(0) = -3(0)^2 + 12 = 12 The y-intercept is (0, 12). Step 3: Determine the x-intercepts. Set f(x) = 0: 0 = -3x^2 + 12 3x^2 = 12 x^2 = (12)/(3) x^2 = 4 x = ± sqrt(4) x = ± 2 The x-intercepts are (-2, 0) and (2, 0). Sketch description: The graph is a parabola opening downwards. It passes through the x-axis at (-2, 0) and (2, 0). It passes through the y-axis at (0, 12). The turning point (which is a maximum) is at (0, 12). 4 (b) Does f(x) have a maximum or minimum value? Determine this value. Since the parabola opens downwards (a = -3 < 0), it has a maximum value. The maximum value is the y-coordinate of the turning point. Maximum value: y = 12 **4 (c) What is the domain and range of f(