1. Write the function $f(x) = 2x^2 - 7x - 10$, where $x \in \mathbb{R}$, in the form $a(x + h)^2 + k$, where $a, h, \text{ and } k \in \mathbb{Q}$. 2. Hence, write the minimum point of $f$. 3. Explain why $f$ must have two real roots. 4. Write the roots of $f(x) = 0$ in the form $p \pm \sqrt{q}$, where $p \text{ and } q \in \mathbb{Q}$.

Photo AI

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

Question 1

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

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

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

Step 1

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

96%

114 rated

Answer

To write the function in the form $a(x + h)^2 + k$ , we can complete the square:

Factor out the coefficient of $x^2$ , which is $2$ :

$f(x) = 2(x^2 - \frac{7}{2}x) - 10$
Find the term to complete the square. Take half the coefficient of $x$ (which is $\frac{-7}{2}$ ), square it $(\frac{-7}{4})^2 = \frac{49}{16}$ , and add and subtract this inside the parentheses:

$f(x) = 2\left(x^2 - \frac{7}{2}x + \frac{49}{16} - \frac{49}{16}\right) - 10$
Simplify:

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

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

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

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

Thus, the function is in the required form: $f(x) = 2(x - \frac{7}{4})^2 - \frac{129}{8}$ .

Step 2

Hence, write the minimum point of $f$

99%

104 rated

Answer

The minimum point occurs at the vertex of the parabola, which is given by the coordinates $(h, k)$ from the completed square form. Here, $h = \frac{7}{4}$ and $k = -\frac{129}{8}$ . Thus, the minimum point of $f$ is:

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

Step 3

Explain why $f$ must have two real roots

96%

101 rated

Answer

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

$D = b^2 - 4ac$

In this case, $a = 2$ , $b = -7$ , and $c = -10$ .

Calculating the discriminant:

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

Since $D > 0$ , the quadratic equation must have two real roots.

Step 4

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

98%

120 rated

Answer

To find the roots of the equation $f(x) = 0$ , we set:

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

Dividing through by 2:

$(x^2 - \frac{7}{2}x - 5 = 0)$

Using the quadratic formula, $x = \frac{-b \pm \sqrt{D}}{2a}$ :

Substituting the values:

Discriminant $D = 129$
Coefficients: $a = 1, b = -\frac{7}{2}$

Thus:

$x = \frac{\frac{7}{2} \pm \sqrt{129}}{2}$

Simplifying, we get:

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

Thus, the roots can be expressed as:

$x = \frac{7}{4} \pm \sqrt{\frac{129}{16}}$

or in the form:

$p \pm \sqrt{q}$ where $p = \frac{7}{4}$ and $q = \frac{129}{16}$ .

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

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

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

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 $p \pm \sqrt{q}$

Join the Leaving Cert students using SimpleStudy...