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8. (a) Sketch the graph of $y = 3^x$, $x \in \mathbb{R}$ showing the coordinates of any points at which the graph crosses the axes - Edexcel - A-Level Maths Pure - Question 10 - 2014 - Paper 1

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8.-(a)-Sketch-the-graph-of--------$y-=-3^x$,-$x-\in-\mathbb{R}$--------showing-the-coordinates-of-any-points-at-which-the-graph-crosses-the-axes-Edexcel-A-Level Maths Pure-Question 10-2014-Paper 1.png

8. (a) Sketch the graph of $y = 3^x$, $x \in \mathbb{R}$ showing the coordinates of any points at which the graph crosses the axes. (b) Use algebra... show full transcript

Worked Solution & Example Answer:8. (a) Sketch the graph of $y = 3^x$, $x \in \mathbb{R}$ showing the coordinates of any points at which the graph crosses the axes - Edexcel - A-Level Maths Pure - Question 10 - 2014 - Paper 1

Step 1

Sketch the graph of $y = 3^x$

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Answer

To sketch the graph of the function y=3xy = 3^x, we will first identify key characteristics and points:

  1. Intercept with the y-axis: When x=0x = 0,

    y=30=1y = 3^0 = 1.

    Therefore, the graph crosses the y-axis at the point (0,1)(0, 1).

  2. Intercept with the x-axis: For y=3xy = 3^x to equal 0,

    there are no points where this is true since an exponential function never crosses the x-axis.

  3. Behavior of the graph:

    • As xx \to -\infty, y0y \to 0.
    • As xx \to \infty, yy \to \infty.
  4. Shape of the curve: The graph is increasing for all xRx \in \mathbb{R}, has no turning points, and approaches the x-axis but never touches it.

In summary, the graph passes through the point (0,1)(0, 1) and approaches the x-axis as xx decreases, indicating that it does not touch the x-axis.

Step 2

Use algebra to solve the equation $3^{2x} - 9(3^x) + 18 = 0$

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Answer

To solve the equation 32x9(3x)+18=03^{2x} - 9(3^x) + 18 = 0, we start by substituting:

Let k=3xk = 3^x. Then, we can rewrite the equation:

k29k+18=0k^2 - 9k + 18 = 0.

Next, we will factor the quadratic equation:

(k6)(k3)=0.(k - 6)(k - 3) = 0.

This gives us two solutions:

  1. k=6k = 6 leads to 3x=63^x = 6.

    Taking the logarithm of both sides, we have:

    x=log36=log6log31.63.x = \log_3 6 = \frac{\log 6}{\log 3} \approx 1.63.

  2. k=3k = 3 leads to 3x=33^x = 3.

    Similarly, we find:

    x=1.x = 1.

Finally, our answers are x1.63x \approx 1.63 and x=1x = 1, with 1.631.63 rounded to two decimal places where appropriate.

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