Write the function $f(x) = 2x^2 - 7x - 10$, where $x \in \mathbb{R}$, in the form $\alpha(x + h)^2 + k$, where $\alpha, h,$ and $k \in \mathbb{Q}$. Hence, write the minimum point of $f$. (i) Explain why $f$ must have two real roots. (ii) Write the roots of $f(x) = 0$ in the form $p \pm \sqrt{q}$, where $p$ and $q \in \mathbb{Q}$.

Leaving Cert Mathematics Algebra: Write the function $f(x) = 2x^2

Write the function $f(x) = 2x^2 - 7x - 10$, where $x \in \mathbb{R}$, in the form $\alpha(x + h)^2 + k$, where $\alpha, h,$ and $k \in \mathbb{Q}$ - Leaving Cert Mathematics - Question 1 - 2017

Question 1

$Write-the-function-$f(x)-=-2x^2---7x---10$,-where-$x-\in-\mathbb{R}$,-in-the-form-$\alpha(x-+-h)^2-+-k$,-where-$\alpha,-h,$-and-$k-\in-\mathbb{Q}$-Leaving Cert Mathematics-Question 1-2017.png$

Write the function $f(x) = 2x^2 - 7x - 10$, where $x \in \mathbb{R}$, in the form $\alpha(x + h)^2 + k$, where $\alpha, h,$ and $k \in \mathbb{Q}$. Hence, write t... show full transcript

Worked Solution & Example Answer:Write the function $f(x) = 2x^2 - 7x - 10$, where $x \in \mathbb{R}$, in the form $\alpha(x + h)^2 + k$, where $\alpha, h,$ and $k \in \mathbb{Q}$ - Leaving Cert Mathematics - Question 1 - 2017

Step 1

Write the function $f(x) = 2x^2 - 7x - 10$ in the form $\alpha(x + h)^2 + k$

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Answer

To rewrite the function in the completed square form, we start with:

$f(x) = 2x^2 - 7x - 10$

First, factor out the 2 from the quadratic terms:

$f(x) = 2(x^2 - \frac{7}{2}x) - 10$

Next, complete the square for the expression inside the parentheses. We take half of the coefficient of $x$ , square it, and add and subtract it inside:

$f(x) = 2 \left( x^2 - \frac{7}{2}x + \left(\frac{7}{4}\right)^2 - \left(\frac{7}{4}\right)^2 \right) - 10$

Simplifying further:

$= 2 \left( \left( x - \frac{7}{4} \right)^2 - \frac{49}{16} \right) - 10$

$= 2 \left( x - \frac{7}{4} \right)^2 - \frac{49}{8} - 10$

Now, convert -10 into eighths:

$= 2 \left( x - \frac{7}{4} \right)^2 - \frac{49}{8} - \frac{80}{8}$

Finally, we have:

$f(x) = 2 \left( x - \frac{7}{4} \right)^2 - \frac{129}{8}$

Step 2

Hence, write the minimum point of $f$

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Answer

The minimum point of the function occurs at the vertex of the parabola, which is given by the coordinates $\left( \frac{7}{4}, -\frac{129}{8} \right)$ . Therefore, the minimum point of $f$ is:

$\left( \frac{7}{4}, -\frac{129}{8} \right)$

Step 3

Explain why $f$ must have two real roots

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Answer

For the function $f(x) = 2x^2 - 7x - 10$ to have two real roots, the discriminant of the quadratic must be greater than zero. The discriminant is given by:

$D = b^2 - 4ac$

where $a = 2$ , $b = -7$ , and $c = -10$ . Thus:

$D = (-7)^2 - 4(2)(-10) = 49 + 80 = 129 > 0$

Since the discriminant is positive, $f$ has two distinct real roots.

Step 4

Write the roots of $f(x) = 0$ in the form $p \pm \sqrt{q}$

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Answer

To find the roots of the quadratic equation, we set:

$2x^2 - 7x - 10 = 0$

Using the quadratic formula:

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

Substituting the values of $a$ , $b$ , and $c$ :

$x = \frac{7 \pm \sqrt{129}}{4}$

Thus, the roots can be written in the form:

$x = \frac{7}{4} \pm \frac{\sqrt{129}}{4}$

Write the function $f(x) = 2x^2 - 7x - 10$, where $x \in \mathbb{R}$, in the form $\alpha(x + h)^2 + k$, where $\alpha, h,$ and $k \in \mathbb{Q}$ - Leaving Cert Mathematics - Question 1 - 2017

Worked Solution & Example Answer:Write the function $f(x) = 2x^2 - 7x - 10$, where $x \in \mathbb{R}$, in the form $\alpha(x + h)^2 + k$, where $\alpha, h,$ and $k \in \mathbb{Q}$ - Leaving Cert Mathematics - Question 1 - 2017

Write the function $f(x) = 2x^2 - 7x - 10$ in the form $\alpha(x + h)^2 + k$

Hence, write the minimum point of $f$

Explain why $f$ must have two real roots

Write the roots of $f(x) = 0$ in the form $p \pm \sqrt{q}$

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