Find expert answers and community-driven knowledge on IDNLearn.com. Whether it's a simple query or a complex problem, our community has the answers you need.
Sagot :
To find the y-intercept and the minimum of the quadratic function [tex]\( y = x^2 + 4x - 5 \)[/tex], let's go through the steps involved.
### Finding the y-intercept
The y-intercept of a function is the point where the graph of the function crosses the y-axis. This occurs when [tex]\( x = 0 \)[/tex].
1. Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ y = (0)^2 + 4(0) - 5 = -5 \][/tex]
So, the y-intercept of the function is [tex]\( (0, -5) \)[/tex].
### Finding the minimum of the function
A quadratic function of the form [tex]\( y = ax^2 + bx + c \)[/tex] has a parabolic shape. If the coefficient [tex]\( a \)[/tex] is positive, the parabola opens upwards and the vertex represents the minimum point of the function.
The vertex (h, k) of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] can be found using:
[tex]\[ h = -\frac{b}{2a} \][/tex]
For our function [tex]\( y = x^2 + 4x - 5 \)[/tex]:
1. The coefficients are [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -5 \)[/tex].
2. Calculate [tex]\( h \)[/tex] (the x-coordinate of the vertex):
[tex]\[ h = -\frac{b}{2a} = -\frac{4}{2 \cdot 1} = -2 \][/tex]
3. Substitute [tex]\( h = -2 \)[/tex] back into the function to find the y-coordinate ([tex]\( k \)[/tex]) of the vertex:
[tex]\[ y = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9 \][/tex]
So, the vertex of the function, which represents the minimum point, is [tex]\((-2, -9)\)[/tex].
### Summary
- The y-intercept of the function [tex]\( y = x^2 + 4x - 5 \)[/tex] is [tex]\((0, -5)\)[/tex].
- The minimum point of the function is [tex]\((-2, -9)\)[/tex].
### Finding the y-intercept
The y-intercept of a function is the point where the graph of the function crosses the y-axis. This occurs when [tex]\( x = 0 \)[/tex].
1. Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ y = (0)^2 + 4(0) - 5 = -5 \][/tex]
So, the y-intercept of the function is [tex]\( (0, -5) \)[/tex].
### Finding the minimum of the function
A quadratic function of the form [tex]\( y = ax^2 + bx + c \)[/tex] has a parabolic shape. If the coefficient [tex]\( a \)[/tex] is positive, the parabola opens upwards and the vertex represents the minimum point of the function.
The vertex (h, k) of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] can be found using:
[tex]\[ h = -\frac{b}{2a} \][/tex]
For our function [tex]\( y = x^2 + 4x - 5 \)[/tex]:
1. The coefficients are [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -5 \)[/tex].
2. Calculate [tex]\( h \)[/tex] (the x-coordinate of the vertex):
[tex]\[ h = -\frac{b}{2a} = -\frac{4}{2 \cdot 1} = -2 \][/tex]
3. Substitute [tex]\( h = -2 \)[/tex] back into the function to find the y-coordinate ([tex]\( k \)[/tex]) of the vertex:
[tex]\[ y = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9 \][/tex]
So, the vertex of the function, which represents the minimum point, is [tex]\((-2, -9)\)[/tex].
### Summary
- The y-intercept of the function [tex]\( y = x^2 + 4x - 5 \)[/tex] is [tex]\((0, -5)\)[/tex].
- The minimum point of the function is [tex]\((-2, -9)\)[/tex].
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for choosing IDNLearn.com. We’re here to provide reliable answers, so please visit us again for more solutions.