Answer :
Sure, let's go through the detailed solution step-by-step for the given expression [tex]\( y = x^2 + 4x - 5 \)[/tex].
### Step 1: Understand the Expression
We are given a quadratic equation in the form [tex]\( y = ax^2 + bx + c \)[/tex]. For our specific equation:
- [tex]\( a = 1 \)[/tex] (coefficient of [tex]\( x^2 \)[/tex])
- [tex]\( b = 4 \)[/tex] (coefficient of [tex]\( x \)[/tex])
- [tex]\( c = -5 \)[/tex] (constant term)
### Step 2: Recognize the Structure
This is a quadratic equation, which is a polynomial of degree 2. The graph of this equation will be a parabola. Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a \)[/tex]) is positive, the parabola will open upwards.
### Step 3: Find the Vertex
The vertex form of a quadratic equation is [tex]\( y = a(x-h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex.
To find the vertex [tex]\( (h, k) \)[/tex] of the equation [tex]\( y = x^2 + 4x - 5 \)[/tex]:
1. Calculate [tex]\( h \)[/tex] using the formula [tex]\( h = -\frac{b}{2a} \)[/tex]:
[tex]\[ h = -\frac{4}{2 \times 1} = -\frac{4}{2} = -2 \][/tex]
2. Calculate [tex]\( k \)[/tex] by substituting [tex]\( h \)[/tex] back into the original equation:
[tex]\[ k = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9 \][/tex]
So, the vertex of the parabola is at [tex]\( (-2, -9) \)[/tex].
### Step 4: Find the Y-Intercept
The y-intercept is the point where the graph intersects the y-axis. This occurs when [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 0^2 + 4(0) - 5 = -5 \][/tex]
So, the y-intercept is [tex]\( (0, -5) \)[/tex].
### Step 5: Find the X-Intercepts (Roots)
To find the x-intercepts (where the graph intersects the x-axis), we set [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = x^2 + 4x - 5 \][/tex]
We solve the quadratic equation:
[tex]\[ x^2 + 4x - 5 = 0 \][/tex]
Factor the quadratic expression:
[tex]\[ (x + 5)(x - 1) = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x + 5 = 0 \quad \implies \quad x = -5 \][/tex]
[tex]\[ x - 1 = 0 \quad \implies \quad x = 1 \][/tex]
So, the x-intercepts are [tex]\( x = -5 \)[/tex] and [tex]\( x = 1 \)[/tex], or the points [tex]\( (-5, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
### Step 6: Sketch the Graph
Using the information we've calculated:
- The parabola opens upwards.
- The vertex is at [tex]\( (-2, -9) \)[/tex].
- The y-intercept is [tex]\( (0, -5) \)[/tex].
- The x-intercepts are [tex]\( (-5, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
These points and the identified behavior of the parabola allow us to sketch the general shape of the graph.
### Summary
The equation [tex]\( y = x^2 + 4x - 5 \)[/tex] represents a parabola that opens upwards with:
- Vertex at [tex]\( (-2, -9) \)[/tex]
- Y-intercept at [tex]\( (0, -5) \)[/tex]
- X-intercepts at [tex]\( (-5, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex]
We have taken into account all the critical points and characteristics needed to understand and graph this quadratic function.
### Step 1: Understand the Expression
We are given a quadratic equation in the form [tex]\( y = ax^2 + bx + c \)[/tex]. For our specific equation:
- [tex]\( a = 1 \)[/tex] (coefficient of [tex]\( x^2 \)[/tex])
- [tex]\( b = 4 \)[/tex] (coefficient of [tex]\( x \)[/tex])
- [tex]\( c = -5 \)[/tex] (constant term)
### Step 2: Recognize the Structure
This is a quadratic equation, which is a polynomial of degree 2. The graph of this equation will be a parabola. Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a \)[/tex]) is positive, the parabola will open upwards.
### Step 3: Find the Vertex
The vertex form of a quadratic equation is [tex]\( y = a(x-h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex.
To find the vertex [tex]\( (h, k) \)[/tex] of the equation [tex]\( y = x^2 + 4x - 5 \)[/tex]:
1. Calculate [tex]\( h \)[/tex] using the formula [tex]\( h = -\frac{b}{2a} \)[/tex]:
[tex]\[ h = -\frac{4}{2 \times 1} = -\frac{4}{2} = -2 \][/tex]
2. Calculate [tex]\( k \)[/tex] by substituting [tex]\( h \)[/tex] back into the original equation:
[tex]\[ k = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9 \][/tex]
So, the vertex of the parabola is at [tex]\( (-2, -9) \)[/tex].
### Step 4: Find the Y-Intercept
The y-intercept is the point where the graph intersects the y-axis. This occurs when [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 0^2 + 4(0) - 5 = -5 \][/tex]
So, the y-intercept is [tex]\( (0, -5) \)[/tex].
### Step 5: Find the X-Intercepts (Roots)
To find the x-intercepts (where the graph intersects the x-axis), we set [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = x^2 + 4x - 5 \][/tex]
We solve the quadratic equation:
[tex]\[ x^2 + 4x - 5 = 0 \][/tex]
Factor the quadratic expression:
[tex]\[ (x + 5)(x - 1) = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x + 5 = 0 \quad \implies \quad x = -5 \][/tex]
[tex]\[ x - 1 = 0 \quad \implies \quad x = 1 \][/tex]
So, the x-intercepts are [tex]\( x = -5 \)[/tex] and [tex]\( x = 1 \)[/tex], or the points [tex]\( (-5, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
### Step 6: Sketch the Graph
Using the information we've calculated:
- The parabola opens upwards.
- The vertex is at [tex]\( (-2, -9) \)[/tex].
- The y-intercept is [tex]\( (0, -5) \)[/tex].
- The x-intercepts are [tex]\( (-5, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
These points and the identified behavior of the parabola allow us to sketch the general shape of the graph.
### Summary
The equation [tex]\( y = x^2 + 4x - 5 \)[/tex] represents a parabola that opens upwards with:
- Vertex at [tex]\( (-2, -9) \)[/tex]
- Y-intercept at [tex]\( (0, -5) \)[/tex]
- X-intercepts at [tex]\( (-5, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex]
We have taken into account all the critical points and characteristics needed to understand and graph this quadratic function.
