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Solve the equation $-x^2 + 6x - 4 = 0$ - Leaving Cert Mathematics - Question 4 - 2015

Question 4

Solve the equation $-x^2 + 6x - 4 = 0$. Give each solution correct to one decimal place. Find the co-ordinates of the turning point of the function $f(x) = -x^2 + ... show full transcript

Worked Solution & Example Answer:Solve the equation $-x^2 + 6x - 4 = 0$ - Leaving Cert Mathematics - Question 4 - 2015

Step 1

Solve the equation $-x^2 + 6x - 4 = 0$

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Answer

To solve the equation, rearrange it into standard quadratic form:

$-x^2 + 6x - 4 = 0$

Next, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = -1$ , $b = 6$ , and $c = -4$ .

Calculating the discriminant:

$b^2 - 4ac = 6^2 - 4(-1)(-4) = 36 - 16 = 20$

Now substituting back into the quadratic formula:

$x = \frac{-6 \pm \sqrt{20}}{2(-1)}$

This simplifies to:

$x = \frac{6 \pm 2\sqrt{5}}{2}$

So we have:

$x = 3 \pm \sqrt{5}$

Calculating the two approximate solutions gives:

$x_1 = 3 + \sqrt{5} \approx 5.2$ $x_2 = 3 - \sqrt{5} \approx 0.8$

Thus, the solutions are $x \approx 5.2$ and $x \approx 0.8$ .

Step 2

Find the co-ordinates of the turning point of the function $f(x) = -x^2 + 6x - 4, \, x \in R$

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Answer

To find the turning point, we first take the derivative of the function:

$f'(x) = -2x + 6$

Setting the derivative to 0 to find critical points:

$-2x + 6 = 0 \Rightarrow x = 3$

Next, we evaluate the function at this point to find the y-coordinate:

$f(3) = -3^2 + 6(3) - 4 = -9 + 18 - 4 = 5$

Thus, the coordinates of the turning point are $(3, 5)$ .

Step 3

Use your answers to parts (a) and (b) above to sketch the curve $y = f(x)$

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Answer

To sketch the curve $y = -x^2 + 6x - 4$ :

Identify the x-intercepts found in part (a): $x \approx 5.2$ and $x \approx 0.8$ .
Mark the turning point $(3, 5)$ , which is the vertex of the parabola.
Since the leading coefficient is negative, the parabola opens downwards. Therefore, draw a smooth curve peaking at the turning point and intersecting the x-axis at the calculated intercepts.
Ensure to label your axes with appropriate scales along both x and y to provide a clear representation of the curve.

Solve the equation $-x^2 + 6x - 4 = 0$ - Leaving Cert Mathematics - Question 4 - 2015

Worked Solution & Example Answer:Solve the equation $-x^2 + 6x - 4 = 0$ - Leaving Cert Mathematics - Question 4 - 2015

Solve the equation $-x^2 + 6x - 4 = 0$

Find the co-ordinates of the turning point of the function $f(x) = -x^2 + 6x - 4, \, x \in R$

Use your answers to parts (a) and (b) above to sketch the curve $y = f(x)$

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