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Sagot :
Sure, let's go through each part of the question step-by-step.
### 1. Finding the y-intercept
The y-intercept of a function is the value of [tex]\( f(x) \)[/tex] when [tex]\( x = 0 \)[/tex].
Plugging [tex]\( x = 0 \)[/tex] into the function [tex]\( f(x) = -5x^2 + 10x - 3 \)[/tex]:
[tex]\[ f(0) = -5(0)^2 + 10(0) - 3 = -3 \][/tex]
So, the y-intercept is:
[tex]\[ \boxed{-3} \][/tex]
### 2. Finding the x-intercepts
The x-intercepts occur where the function [tex]\( f(x) = 0 \)[/tex]. In other words, we need to solve the quadratic equation:
[tex]\[ -5x^2 + 10x - 3 = 0 \][/tex]
This is a standard quadratic equation in the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. Here, [tex]\( a = -5 \)[/tex], [tex]\( b = 10 \)[/tex], and [tex]\( c = -3 \)[/tex].
The solutions to this quadratic equation can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, we calculate the discriminant:
[tex]\[ b^2 - 4ac = 10^2 - 4(-5)(-3) = 100 - 60 = 40 \][/tex]
Next, we take the square root of the discriminant:
[tex]\[ \sqrt{40} \approx 6.3246 \][/tex]
Now, substitute into the quadratic formula:
[tex]\[ x = \frac{-10 \pm 6.3246}{-10} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{10 + 6.3246}{10} = 1.6325 \][/tex]
[tex]\[ x_2 = \frac{10 - 6.3246}{10} = 0.3675 \][/tex]
So, the x-intercepts are approximately:
[tex]\[ \boxed{0.3675 \text{ and } 1.6325} \][/tex]
### 3. Finding the Turning Point (Vertex)
The turning point of a parabola given by [tex]\( ax^2 + bx + c \)[/tex] is at:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -5 \)[/tex] and [tex]\( b = 10 \)[/tex]:
[tex]\[ x = -\frac{10}{2(-5)} = 1 \][/tex]
Now we need to find [tex]\( f(1) \)[/tex] to get the y-coordinate of the vertex:
[tex]\[ f(1) = -5(1)^2 + 10(1) - 3 = -5 + 10 - 3 = 2 \][/tex]
So, the turning point (vertex) is at:
[tex]\[ \boxed{(1, 2)} \][/tex]
Summary:
- The y-intercept is [tex]\( -3 \)[/tex].
- The x-intercepts are approximately [tex]\( 0.3675 \)[/tex] and [tex]\( 1.6325 \)[/tex].
- The turning point is at [tex]\( (1, 2) \)[/tex].
### 1. Finding the y-intercept
The y-intercept of a function is the value of [tex]\( f(x) \)[/tex] when [tex]\( x = 0 \)[/tex].
Plugging [tex]\( x = 0 \)[/tex] into the function [tex]\( f(x) = -5x^2 + 10x - 3 \)[/tex]:
[tex]\[ f(0) = -5(0)^2 + 10(0) - 3 = -3 \][/tex]
So, the y-intercept is:
[tex]\[ \boxed{-3} \][/tex]
### 2. Finding the x-intercepts
The x-intercepts occur where the function [tex]\( f(x) = 0 \)[/tex]. In other words, we need to solve the quadratic equation:
[tex]\[ -5x^2 + 10x - 3 = 0 \][/tex]
This is a standard quadratic equation in the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. Here, [tex]\( a = -5 \)[/tex], [tex]\( b = 10 \)[/tex], and [tex]\( c = -3 \)[/tex].
The solutions to this quadratic equation can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, we calculate the discriminant:
[tex]\[ b^2 - 4ac = 10^2 - 4(-5)(-3) = 100 - 60 = 40 \][/tex]
Next, we take the square root of the discriminant:
[tex]\[ \sqrt{40} \approx 6.3246 \][/tex]
Now, substitute into the quadratic formula:
[tex]\[ x = \frac{-10 \pm 6.3246}{-10} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{10 + 6.3246}{10} = 1.6325 \][/tex]
[tex]\[ x_2 = \frac{10 - 6.3246}{10} = 0.3675 \][/tex]
So, the x-intercepts are approximately:
[tex]\[ \boxed{0.3675 \text{ and } 1.6325} \][/tex]
### 3. Finding the Turning Point (Vertex)
The turning point of a parabola given by [tex]\( ax^2 + bx + c \)[/tex] is at:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -5 \)[/tex] and [tex]\( b = 10 \)[/tex]:
[tex]\[ x = -\frac{10}{2(-5)} = 1 \][/tex]
Now we need to find [tex]\( f(1) \)[/tex] to get the y-coordinate of the vertex:
[tex]\[ f(1) = -5(1)^2 + 10(1) - 3 = -5 + 10 - 3 = 2 \][/tex]
So, the turning point (vertex) is at:
[tex]\[ \boxed{(1, 2)} \][/tex]
Summary:
- The y-intercept is [tex]\( -3 \)[/tex].
- The x-intercepts are approximately [tex]\( 0.3675 \)[/tex] and [tex]\( 1.6325 \)[/tex].
- The turning point is at [tex]\( (1, 2) \)[/tex].
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