Answer :
To solve the problem of graphing the parabola given by the equation [tex]\( y = x^2 + 11x + 28 \)[/tex] and finding its vertex, axis of symmetry, x-intercepts, and y-intercept, we will proceed through the following steps.
### 1. Finding the Vertex
The vertex of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] occurs at the point [tex]\( \left( -\frac{b}{2a}, y \right) \)[/tex].
For the equation [tex]\( y = x^2 + 11x + 28 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 11 \)[/tex]
- [tex]\( c = 28 \)[/tex]
Using the formula for the x-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} = -\frac{11}{2 \cdot 1} = -\frac{11}{2} = -5.5 \][/tex]
To find the y-coordinate of the vertex, we substitute [tex]\( x = -5.5 \)[/tex] back into the equation [tex]\( y = x^2 + 11x + 28 \)[/tex]:
[tex]\[ y = (-5.5)^2 + 11(-5.5) + 28 \][/tex]
[tex]\[ y = 30.25 - 60.5 + 28 \][/tex]
[tex]\[ y = 30.25 - 32.5 \][/tex]
[tex]\[ y = -2.25 \][/tex]
Thus, the vertex is [tex]\( \left( -5.5, -2.25 \right) \)[/tex].
### 2. Axis of Symmetry
The axis of symmetry is the vertical line that passes through the vertex of the parabola. Since the x-coordinate of the vertex is [tex]\( -5.5 \)[/tex], the equation of the axis of symmetry is:
[tex]\[ x = -5.5 \][/tex]
### 3. Finding the X-Intercepts
The x-intercepts occur where [tex]\( y = 0 \)[/tex]. Therefore, we solve the quadratic equation:
[tex]\[ x^2 + 11x + 28 = 0 \][/tex]
To find the roots, we can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 1, \, b = 11, \, c = 28 \][/tex]
The discriminant ([tex]\( b^2 - 4ac \)[/tex]) is:
[tex]\[ 11^2 - 4 \cdot 1 \cdot 28 = 121 - 112 = 9 \][/tex]
Taking the square root of the discriminant:
[tex]\[ \sqrt{9} = 3 \][/tex]
Thus, the roots are:
[tex]\[ x = \frac{-11 \pm 3}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-11 + 3}{2} = \frac{-8}{2} = -4 \][/tex]
[tex]\[ x = \frac{-11 - 3}{2} = \frac{-14}{2} = -7 \][/tex]
Therefore, the x-intercepts are [tex]\( (-4, 0) \)[/tex] and [tex]\( (-7, 0) \)[/tex].
### 4. Finding the Y-Intercept
The y-intercept occurs where [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 0^2 + 11 \cdot 0 + 28 \][/tex]
[tex]\[ y = 28 \][/tex]
Thus, the y-intercept is [tex]\( (0, 28) \)[/tex].
### Summary
1. Vertex: [tex]\( \left( -5.5, -2.25 \right) \)[/tex]
2. Axis of Symmetry: [tex]\( x = -5.5 \)[/tex]
3. X-Intercepts: [tex]\( (-4, 0) \)[/tex] and [tex]\( (-7, 0) \)[/tex]
4. Y-Intercept: [tex]\( (0, 28) \)[/tex]
When graphing the parabola, be sure to accurately plot these points and the symmetric nature around the axis of symmetry [tex]\( x = -5.5 \)[/tex].
### 1. Finding the Vertex
The vertex of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] occurs at the point [tex]\( \left( -\frac{b}{2a}, y \right) \)[/tex].
For the equation [tex]\( y = x^2 + 11x + 28 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 11 \)[/tex]
- [tex]\( c = 28 \)[/tex]
Using the formula for the x-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} = -\frac{11}{2 \cdot 1} = -\frac{11}{2} = -5.5 \][/tex]
To find the y-coordinate of the vertex, we substitute [tex]\( x = -5.5 \)[/tex] back into the equation [tex]\( y = x^2 + 11x + 28 \)[/tex]:
[tex]\[ y = (-5.5)^2 + 11(-5.5) + 28 \][/tex]
[tex]\[ y = 30.25 - 60.5 + 28 \][/tex]
[tex]\[ y = 30.25 - 32.5 \][/tex]
[tex]\[ y = -2.25 \][/tex]
Thus, the vertex is [tex]\( \left( -5.5, -2.25 \right) \)[/tex].
### 2. Axis of Symmetry
The axis of symmetry is the vertical line that passes through the vertex of the parabola. Since the x-coordinate of the vertex is [tex]\( -5.5 \)[/tex], the equation of the axis of symmetry is:
[tex]\[ x = -5.5 \][/tex]
### 3. Finding the X-Intercepts
The x-intercepts occur where [tex]\( y = 0 \)[/tex]. Therefore, we solve the quadratic equation:
[tex]\[ x^2 + 11x + 28 = 0 \][/tex]
To find the roots, we can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 1, \, b = 11, \, c = 28 \][/tex]
The discriminant ([tex]\( b^2 - 4ac \)[/tex]) is:
[tex]\[ 11^2 - 4 \cdot 1 \cdot 28 = 121 - 112 = 9 \][/tex]
Taking the square root of the discriminant:
[tex]\[ \sqrt{9} = 3 \][/tex]
Thus, the roots are:
[tex]\[ x = \frac{-11 \pm 3}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-11 + 3}{2} = \frac{-8}{2} = -4 \][/tex]
[tex]\[ x = \frac{-11 - 3}{2} = \frac{-14}{2} = -7 \][/tex]
Therefore, the x-intercepts are [tex]\( (-4, 0) \)[/tex] and [tex]\( (-7, 0) \)[/tex].
### 4. Finding the Y-Intercept
The y-intercept occurs where [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 0^2 + 11 \cdot 0 + 28 \][/tex]
[tex]\[ y = 28 \][/tex]
Thus, the y-intercept is [tex]\( (0, 28) \)[/tex].
### Summary
1. Vertex: [tex]\( \left( -5.5, -2.25 \right) \)[/tex]
2. Axis of Symmetry: [tex]\( x = -5.5 \)[/tex]
3. X-Intercepts: [tex]\( (-4, 0) \)[/tex] and [tex]\( (-7, 0) \)[/tex]
4. Y-Intercept: [tex]\( (0, 28) \)[/tex]
When graphing the parabola, be sure to accurately plot these points and the symmetric nature around the axis of symmetry [tex]\( x = -5.5 \)[/tex].