\( \left.\begin{array}{l}5.1 \text { Solve the value of } x \text { and } y \text { using determinants if given: } \\ 5.2\left|\begin{array}{ccc}x & y & 1 \\ 1 & -1 & 1 \\ 13 & 2 & 1\end{array}\right|=0 \\ 5.3\end{array} \begin{array}{ccc}x & y & 1 \\ 3 & 2 & 1 \\ -6 & -4 & 1\end{array} \right\rvert\,=0 \quad \) Solve for \( x \) if \( \sqrt{2^{x}}+\sqrt{2^{x+2}}=24 \) (8) \( \begin{array}{l}\text { Sketch the graph of the parabola }-x^{2}+6 x-8 \text { and clearly indicate the } \\ \text { intercepts with the axes as well as the turning points. } \\ \text { Is the graph }-x^{2}+6 x-8 \text { continuous or discontinuous? }\end{array} \)

We will solve the problem step by step. ────────────────────────────── Step 1. Solve for x and y using determinants. We are given two determinants equal to zero. Recall that if three points A, B, and C (written in homogeneous coordinates as (x, y, 1), (x₁, y₁, 1), (x₂, y₂, 1)) are collinear, then   | x   y   1 |   | x₁  y₁  1 | = 0   | x₂  y₂  1 | represents the condition for collinearity. Thus, each determinant below gives the equation of a straight line through the two fixed points, and (x, y) is a point on that line. ────────────────────────────── 1.1. First Determinant Determinant:   | x   y   1 |   | 1  -1   1 |   | 13  2   1 | = 0 This means (x, y) lies on the line through (1, -1) and (13, 2). Find the slope between (1, -1) and (13, 2):   Slope m = (2 – (–1))⁄(13 – 1) = 3⁄12 = ¼. Using point-slope form (with point (1, -1)):   y – (–1) = (1/4)(x – 1)   ⇒ y + 1 = (1/4)(x – 1)   ⇒ y = (1/4)(x – 1) – 1 Simplify:   y = x/4 – 1/4 – 1 = x/4 – 5/4. So the first line is:   4y = x – 5            (Equation 1) ────────────────────────────── 1.2. Second Determinant Determinant:   | x   y   1 |   | 3  2   1 |   | –6 –4   1 | = 0 This signifies that (x, y) is collinear with (3, 2) and (–6, –4). Find the slope between (3, 2) and (–6, –4):   m = [–4 – 2]⁄[–6 – 3] = (–6)⁄(–9) = 2/3. Using point-slope form (with point (3, 2)):   y – 2 = (2/3)(x – 3)   ⇒ y = (2/3)(x – 3) + 2 Simplify:   y = (2/3)x – 2 + 2 = (2/3)x. Thus the second line is:   y = (2/3)x            (Equation 2) ────────────────────────────── 1.3. Find the Intersection (x, y) To find the values of x and y that satisfy both equations, equate the two expressions for y: From Equation 1:  y = x/4 – 5/4 From Equation 2:  y = (2/3)x Set them equal:   (2/3)x = x/4 – 5/4 Solve for x:   Multiply both sides by 12 (the least common multiple of 3 and 4) to eliminate fractions:     12·(2/3)x = 12·(x/4) – 12·(5/4)     8x = 3x – 15 Subtract 3x from both sides:   8x – 3x = –15   5x = –15   x = –3 Then substitute x = –3 into Equation 2:   y = (2/3)(–3) = –2 Thus, the solution is:   (x, y) = (–3, –2). ────────────────────────────── Step 2. Solve for x in the equation   √(2^x) + √(2^(x+2)) = 24 First, express the square roots in exponential form:   √(2^x) = 2^(x/2)   √(2^(x+2)) = 2^((x+2)/2) = 2^(x/2) · 2^(2/2) = 2^(x/2) · 2 So the equation becomes:   2^(x/2) + 2 · 2^(x/2) = 24 Combine like terms:   (1 + 2) · 2^(x/2) = 24   3 · 2^(x/2) = 24 Solve for 2^(x/2):   2^(x/2) = 24/3 = 8 Recall that 8 = 2³, so:   2^(x/2) = 2³ Since the bases are equal, equate the exponents:   x/2 = 3 ⇒ x = 6. ────────────────────────────── Step 3. Sketch the graph of the parabola   y = –x² + 6x – 8 We will identify key features: intercepts and the turning point (vertex). 3.1. x-intercepts Set y = 0:   –x² + 6x – 8 = 0 Multiply both sides by –1:   x² – 6x + 8 = 0 Factor the quadratic:   x² – 6x + 8 = (x – 2)(x – 4) = 0 Thus, x-intercepts are:   x = 2 and x = 4, giving the points (2, 0) and (4, 0). 3.2. y-intercept Find y when x = 0:   y = –0² + 6·0 – 8 = –8 So the y-intercept is (0, –8). 3.3. Vertex (Turning Point) For a quadratic y = ax² + bx + c, the x-coordinate of the vertex is given by:   x = –b/(2a) Here, a = –1 and b = 6:   x = –6/(2·(–1)) = –6/–2 = 3 Now compute the y-coordinate:   y = –(3)² + 6·3 – 8 = –9 + 18 – 8 = 1 So the vertex (turning point) is at (3, 1). 3.4. Graph Characteristics • Because a = –1 (negative), the parabola opens downward. • The axis of symmetry is the vertical line x = 3. When sketching:  • Mark the vertex at (3, 1).  • Plot the x-intercepts at (2, 0) and (4, 0).  • Plot the y-intercept at (0, –8).  • Draw a smooth, downward opening curve through these points. 3.5. Continuity A quadratic function is a polynomial, and polynomials are continuous for all real numbers. Therefore, the graph of y = –x² + 6x – 8 is continuous. ────────────────────────────── Summary of Answers 1. From the determinants, the intersection point of the two lines is:   (x, y) = (–3, –2). 2. Solving √(2^x) + √(2^(x+2)) = 24 gives:   x = 6. 3. The graph of y = –x² + 6x – 8:   • x-intercepts: (2, 0) and (4, 0)   • y-intercept: (0, –8)   • Vertex (turning point): (3, 1)   • The parabola opens downward and is continuous. This completes the solution.