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

Question

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

UpStudy AI · Accepted Answer

We will solve the problem step by step.

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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.

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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)

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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)

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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).

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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.

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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.

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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.
Solutions: 1. The intersection point is (–3, –2). 2. The value of x is 6. 3. The graph of y = –x² + 6x – 8 has x-intercepts at (2, 0) and (4, 0), a y-intercept at (0, –8), a vertex at (3, 1), opens downward, and is continuous.

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