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Sagot :
Certainly! Let's go through each part of the question step by step.
### (a) Express [tex]\( f(x) \)[/tex] in standard form
The standard form (also called vertex form) of a quadratic function is:
[tex]\[ f(x) = a(x - h)^2 + k \][/tex]
where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
The given quadratic function is [tex]\( f(x) = 4x^2 + 8x + 5 \)[/tex].
To convert this to standard form, we can complete the square:
1. Factor out the coefficient of [tex]\( x^2 \)[/tex] from the first two terms:
[tex]\[ f(x) = 4(x^2 + 2x) + 5 \][/tex]
2. Complete the square inside the parentheses:
- Take half of the coefficient of [tex]\( x \)[/tex], which is 2, and square it: [tex]\( \left(\frac{2}{2}\right)^2 = 1 \)[/tex]
- Add and subtract this square inside the parentheses:
[tex]\[ 4(x^2 + 2x + 1 - 1) + 5 = 4((x + 1)^2 - 1) + 5 \][/tex]
3. Distribute the 4 and simplify:
[tex]\[ 4(x + 1)^2 - 4 + 5 = 4(x + 1)^2 + 1 \][/tex]
So the standard (vertex) form of the function is:
[tex]\[ f(x) = 4(x + 1)^2 + 1 \][/tex]
### (b) Find the vertex and [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-intercepts
Vertex:
From the vertex form [tex]\( f(x) = 4(x + 1)^2 + 1 \)[/tex], the vertex [tex]\((h, k)\)[/tex] is [tex]\(( -1, 1 )\)[/tex].
x-intercepts:
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 4x^2 + 8x + 5 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 4 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = 5 \)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 4 \cdot 5}}{2 \cdot 4} = \frac{-8 \pm \sqrt{64 - 80}}{8} = \frac{-8 \pm \sqrt{-16}}{8} = \frac{-8 \pm 4i}{8} = \frac{-2 \pm i}{2} \][/tex]
Since the discriminant ([tex]\( \sqrt{-16} \)[/tex]) is negative, the quadratic equation has no real solutions; hence, there are no real [tex]\( x \)[/tex]-intercepts. Therefore:
[tex]\[ \text{x-intercept: } (x, y) = \text{DNE} \][/tex]
y-intercept:
The [tex]\( y \)[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 4(0)^2 + 8(0) + 5 = 5 \][/tex]
So the [tex]\( y \)[/tex]-intercept is:
[tex]\[ \text{y-intercept: } (x, y) = (0, 5) \][/tex]
### (c) Sketch a graph of [tex]\( f \)[/tex]
To sketch the graph of [tex]\( f(x) = 4x^2 + 8x + 5 \)[/tex]:
1. Plot the vertex at [tex]\( (-1, 1) \)[/tex].
2. Plot the y-intercept at [tex]\( (0, 5) \)[/tex].
3. Determine the direction of the parabola:
The coefficient of [tex]\( x^2 \)[/tex] is 4 (which is positive), so the parabola opens upwards.
Here is a rough sketch of the graph:
```plain
y
↑
10 + . .
| . .
5 + . (0,5)
| .
0 +_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
| . (−1,1)
−5 + . .
| . .
−10+_____________________________.: x
−10 −5 0 5 10
```
The vertex at [tex]\((-1, 1)\)[/tex] represents the minimum point of the parabola, and the [tex]\(y\)[/tex]-intercept at [tex]\( (0, 5) \)[/tex] shows where the graph crosses the [tex]\( y \)[/tex]-axis. The parabola opens upwards indicating that for very large positive or negative values of [tex]\(x\)[/tex], [tex]\(f(x)\)[/tex] will take very large positive values.
### (a) Express [tex]\( f(x) \)[/tex] in standard form
The standard form (also called vertex form) of a quadratic function is:
[tex]\[ f(x) = a(x - h)^2 + k \][/tex]
where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
The given quadratic function is [tex]\( f(x) = 4x^2 + 8x + 5 \)[/tex].
To convert this to standard form, we can complete the square:
1. Factor out the coefficient of [tex]\( x^2 \)[/tex] from the first two terms:
[tex]\[ f(x) = 4(x^2 + 2x) + 5 \][/tex]
2. Complete the square inside the parentheses:
- Take half of the coefficient of [tex]\( x \)[/tex], which is 2, and square it: [tex]\( \left(\frac{2}{2}\right)^2 = 1 \)[/tex]
- Add and subtract this square inside the parentheses:
[tex]\[ 4(x^2 + 2x + 1 - 1) + 5 = 4((x + 1)^2 - 1) + 5 \][/tex]
3. Distribute the 4 and simplify:
[tex]\[ 4(x + 1)^2 - 4 + 5 = 4(x + 1)^2 + 1 \][/tex]
So the standard (vertex) form of the function is:
[tex]\[ f(x) = 4(x + 1)^2 + 1 \][/tex]
### (b) Find the vertex and [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-intercepts
Vertex:
From the vertex form [tex]\( f(x) = 4(x + 1)^2 + 1 \)[/tex], the vertex [tex]\((h, k)\)[/tex] is [tex]\(( -1, 1 )\)[/tex].
x-intercepts:
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 4x^2 + 8x + 5 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 4 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = 5 \)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 4 \cdot 5}}{2 \cdot 4} = \frac{-8 \pm \sqrt{64 - 80}}{8} = \frac{-8 \pm \sqrt{-16}}{8} = \frac{-8 \pm 4i}{8} = \frac{-2 \pm i}{2} \][/tex]
Since the discriminant ([tex]\( \sqrt{-16} \)[/tex]) is negative, the quadratic equation has no real solutions; hence, there are no real [tex]\( x \)[/tex]-intercepts. Therefore:
[tex]\[ \text{x-intercept: } (x, y) = \text{DNE} \][/tex]
y-intercept:
The [tex]\( y \)[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 4(0)^2 + 8(0) + 5 = 5 \][/tex]
So the [tex]\( y \)[/tex]-intercept is:
[tex]\[ \text{y-intercept: } (x, y) = (0, 5) \][/tex]
### (c) Sketch a graph of [tex]\( f \)[/tex]
To sketch the graph of [tex]\( f(x) = 4x^2 + 8x + 5 \)[/tex]:
1. Plot the vertex at [tex]\( (-1, 1) \)[/tex].
2. Plot the y-intercept at [tex]\( (0, 5) \)[/tex].
3. Determine the direction of the parabola:
The coefficient of [tex]\( x^2 \)[/tex] is 4 (which is positive), so the parabola opens upwards.
Here is a rough sketch of the graph:
```plain
y
↑
10 + . .
| . .
5 + . (0,5)
| .
0 +_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
| . (−1,1)
−5 + . .
| . .
−10+_____________________________.: x
−10 −5 0 5 10
```
The vertex at [tex]\((-1, 1)\)[/tex] represents the minimum point of the parabola, and the [tex]\(y\)[/tex]-intercept at [tex]\( (0, 5) \)[/tex] shows where the graph crosses the [tex]\( y \)[/tex]-axis. The parabola opens upwards indicating that for very large positive or negative values of [tex]\(x\)[/tex], [tex]\(f(x)\)[/tex] will take very large positive values.
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