Answer :
Certainly! Let's graph the equation [tex]\( y = -x^2 - 12x - 35 \)[/tex] step-by-step, identifying key points including the vertex and the roots of the parabola.
### Step 1: Identify the shape and direction of the parabola
The equation [tex]\( y = -x^2 - 12x - 35 \)[/tex] represents a quadratic function in the form [tex]\( y = ax^2 + bx + c \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = -35 \)[/tex]. Since the coefficient [tex]\( a \)[/tex] is negative, the parabola opens downwards.
### Step 2: Find the vertex of the parabola
The vertex, for a quadratic equation of the form [tex]\( y = ax^2 + bx + c \)[/tex], can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in the values:
[tex]\[ x = -\frac{-12}{2(-1)} = \frac{12}{-2} = -6 \][/tex]
Now, substitute [tex]\( x = -6 \)[/tex] back into the equation to find the corresponding [tex]\( y \)[/tex]-coordinate:
[tex]\[ y = -(-6)^2 - 12(-6) - 35 \][/tex]
[tex]\[ y = -36 + 72 - 35 \][/tex]
[tex]\[ y = 1 \][/tex]
So, the vertex of the parabola is at [tex]\( (-6, 1) \)[/tex].
### Step 3: Find the roots of the equation (where [tex]\( y = 0 \)[/tex])
To find the roots, we solve the equation [tex]\( -x^2 - 12x - 35 = 0 \)[/tex]. This can be factored or solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting [tex]\( a = -1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = -35 \)[/tex]:
[tex]\[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(-1)(-35)}}{2(-1)} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{144 - 140}}{-2} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{4}}{-2} \][/tex]
[tex]\[ x = \frac{12 \pm 2}{-2} \][/tex]
So, the roots are:
[tex]\[ x = \frac{14}{-2} = -7 \][/tex]
[tex]\[ x = \frac{10}{-2} = -5 \][/tex]
Thus, the roots are at [tex]\( (-7, 0) \)[/tex] and [tex]\( (-5, 0) \)[/tex].
### Step 4: Identify two additional points on the graph
We can choose two [tex]\( x \)[/tex]-values slightly away from the vertex and roots:
- Let [tex]\( x = -8 \)[/tex]:
[tex]\[ y = -(-8)^2 - 12(-8) - 35 \][/tex]
[tex]\[ y = -64 + 96 - 35 \][/tex]
[tex]\[ y = -3 \][/tex]
So, the point is [tex]\( (-8, -3) \)[/tex].
- Let [tex]\( x = -4 \)[/tex]:
[tex]\[ y = -(-4)^2 - 12(-4) - 35 \][/tex]
[tex]\[ y = -16 + 48 - 35 \][/tex]
[tex]\[ y = -3 \][/tex]
So, the point is [tex]\( (-4, -3) \)[/tex].
### Step 5: Plot the points and draw the parabola
We have the following points to plot:
- Vertex: [tex]\( (-6, 1) \)[/tex]
- Roots: [tex]\( (-7, 0) \)[/tex] and [tex]\( (-5, 0) \)[/tex]
- Additional points: [tex]\( (-8, -3) \)[/tex] and [tex]\( (-4, -3) \)[/tex]
Now, plot these points on a graph, and draw a smooth curve through them to represent the parabola.
Graph:
[tex]\[ \begin{array}{c|ccccccc} x & -8 & -7 & -6 & -5 & -4 \\ \hline y & -3 & 0 & 1 & 0 & -3 \\ \end{array} \][/tex]
By plotting these points and joining them smoothly, we obtain the graph of [tex]\( y = -x^2 - 12x - 35 \)[/tex].
### Step 1: Identify the shape and direction of the parabola
The equation [tex]\( y = -x^2 - 12x - 35 \)[/tex] represents a quadratic function in the form [tex]\( y = ax^2 + bx + c \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = -35 \)[/tex]. Since the coefficient [tex]\( a \)[/tex] is negative, the parabola opens downwards.
### Step 2: Find the vertex of the parabola
The vertex, for a quadratic equation of the form [tex]\( y = ax^2 + bx + c \)[/tex], can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in the values:
[tex]\[ x = -\frac{-12}{2(-1)} = \frac{12}{-2} = -6 \][/tex]
Now, substitute [tex]\( x = -6 \)[/tex] back into the equation to find the corresponding [tex]\( y \)[/tex]-coordinate:
[tex]\[ y = -(-6)^2 - 12(-6) - 35 \][/tex]
[tex]\[ y = -36 + 72 - 35 \][/tex]
[tex]\[ y = 1 \][/tex]
So, the vertex of the parabola is at [tex]\( (-6, 1) \)[/tex].
### Step 3: Find the roots of the equation (where [tex]\( y = 0 \)[/tex])
To find the roots, we solve the equation [tex]\( -x^2 - 12x - 35 = 0 \)[/tex]. This can be factored or solved using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting [tex]\( a = -1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = -35 \)[/tex]:
[tex]\[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(-1)(-35)}}{2(-1)} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{144 - 140}}{-2} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{4}}{-2} \][/tex]
[tex]\[ x = \frac{12 \pm 2}{-2} \][/tex]
So, the roots are:
[tex]\[ x = \frac{14}{-2} = -7 \][/tex]
[tex]\[ x = \frac{10}{-2} = -5 \][/tex]
Thus, the roots are at [tex]\( (-7, 0) \)[/tex] and [tex]\( (-5, 0) \)[/tex].
### Step 4: Identify two additional points on the graph
We can choose two [tex]\( x \)[/tex]-values slightly away from the vertex and roots:
- Let [tex]\( x = -8 \)[/tex]:
[tex]\[ y = -(-8)^2 - 12(-8) - 35 \][/tex]
[tex]\[ y = -64 + 96 - 35 \][/tex]
[tex]\[ y = -3 \][/tex]
So, the point is [tex]\( (-8, -3) \)[/tex].
- Let [tex]\( x = -4 \)[/tex]:
[tex]\[ y = -(-4)^2 - 12(-4) - 35 \][/tex]
[tex]\[ y = -16 + 48 - 35 \][/tex]
[tex]\[ y = -3 \][/tex]
So, the point is [tex]\( (-4, -3) \)[/tex].
### Step 5: Plot the points and draw the parabola
We have the following points to plot:
- Vertex: [tex]\( (-6, 1) \)[/tex]
- Roots: [tex]\( (-7, 0) \)[/tex] and [tex]\( (-5, 0) \)[/tex]
- Additional points: [tex]\( (-8, -3) \)[/tex] and [tex]\( (-4, -3) \)[/tex]
Now, plot these points on a graph, and draw a smooth curve through them to represent the parabola.
Graph:
[tex]\[ \begin{array}{c|ccccccc} x & -8 & -7 & -6 & -5 & -4 \\ \hline y & -3 & 0 & 1 & 0 & -3 \\ \end{array} \][/tex]
By plotting these points and joining them smoothly, we obtain the graph of [tex]\( y = -x^2 - 12x - 35 \)[/tex].
