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### Given function: [tex]\( f(x) = x^2 + 7x + 12 \)[/tex]
### Step-by-Step Solution:
#### 5. What is the [tex]\( y \)[/tex]-intercept?
The [tex]\( y \)[/tex]-intercept of a function is the point at which the graph of the function crosses the [tex]\( y \)[/tex]-axis. This occurs when [tex]\( x = 0 \)[/tex].
To find the [tex]\( y \)[/tex]-intercept:
[tex]\[ f(0) = 0^2 + 7(0) + 12 = 12 \][/tex]
So, the [tex]\( y \)[/tex]-intercept is [tex]\( 12 \)[/tex].
#### 6. What are the [tex]\( x \)[/tex]-intercepts?
The [tex]\( x \)[/tex]-intercepts are the points where the graph of the function crosses the [tex]\( x \)[/tex]-axis. These occur when [tex]\( f(x) = 0 \)[/tex].
To find the [tex]\( x \)[/tex]-intercepts, we solve the equation:
[tex]\[ x^2 + 7x + 12 = 0 \][/tex]
Factoring the quadratic:
[tex]\[ (x + 3)(x + 4) = 0 \][/tex]
Setting each factor equal to zero gives us the roots:
[tex]\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \][/tex]
[tex]\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \][/tex]
So, the [tex]\( x \)[/tex]-intercepts are [tex]\( x = -3 \)[/tex] and [tex]\( x = -4 \)[/tex].
#### 7. Where is the turning point?
The turning point of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] occurs at [tex]\( x = -\frac{b}{2a} \)[/tex].
For the quadratic function [tex]\( f(x) = x^2 + 7x + 12 \)[/tex]:
[tex]\[ a = 1, \quad b = 7, \quad c = 12 \][/tex]
To find the [tex]\( x \)[/tex]-coordinate of the turning point:
[tex]\[ x = -\frac{b}{2a} = -\frac{7}{2 \cdot 1} = -\frac{7}{2} = -3.5 \][/tex]
To find the [tex]\( y \)[/tex]-coordinate of the turning point, we substitute [tex]\( x = -3.5 \)[/tex] back into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(-3.5) = (-3.5)^2 + 7(-3.5) + 12 = 12.25 - 24.5 + 12 = -0.25 \][/tex]
So, the turning point is at:
[tex]\[ (-3.5, -0.25) \][/tex]
### Summary of Answers:
1. The [tex]\( y \)[/tex]-intercept is [tex]\( 12 \)[/tex].
2. The [tex]\( x \)[/tex]-intercepts are [tex]\( -3 \)[/tex] and [tex]\( -4 \)[/tex].
3. The turning point is at [tex]\( (-3.5, -0.25) \)[/tex].
### Given function: [tex]\( f(x) = x^2 + 7x + 12 \)[/tex]
### Step-by-Step Solution:
#### 5. What is the [tex]\( y \)[/tex]-intercept?
The [tex]\( y \)[/tex]-intercept of a function is the point at which the graph of the function crosses the [tex]\( y \)[/tex]-axis. This occurs when [tex]\( x = 0 \)[/tex].
To find the [tex]\( y \)[/tex]-intercept:
[tex]\[ f(0) = 0^2 + 7(0) + 12 = 12 \][/tex]
So, the [tex]\( y \)[/tex]-intercept is [tex]\( 12 \)[/tex].
#### 6. What are the [tex]\( x \)[/tex]-intercepts?
The [tex]\( x \)[/tex]-intercepts are the points where the graph of the function crosses the [tex]\( x \)[/tex]-axis. These occur when [tex]\( f(x) = 0 \)[/tex].
To find the [tex]\( x \)[/tex]-intercepts, we solve the equation:
[tex]\[ x^2 + 7x + 12 = 0 \][/tex]
Factoring the quadratic:
[tex]\[ (x + 3)(x + 4) = 0 \][/tex]
Setting each factor equal to zero gives us the roots:
[tex]\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \][/tex]
[tex]\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \][/tex]
So, the [tex]\( x \)[/tex]-intercepts are [tex]\( x = -3 \)[/tex] and [tex]\( x = -4 \)[/tex].
#### 7. Where is the turning point?
The turning point of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] occurs at [tex]\( x = -\frac{b}{2a} \)[/tex].
For the quadratic function [tex]\( f(x) = x^2 + 7x + 12 \)[/tex]:
[tex]\[ a = 1, \quad b = 7, \quad c = 12 \][/tex]
To find the [tex]\( x \)[/tex]-coordinate of the turning point:
[tex]\[ x = -\frac{b}{2a} = -\frac{7}{2 \cdot 1} = -\frac{7}{2} = -3.5 \][/tex]
To find the [tex]\( y \)[/tex]-coordinate of the turning point, we substitute [tex]\( x = -3.5 \)[/tex] back into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(-3.5) = (-3.5)^2 + 7(-3.5) + 12 = 12.25 - 24.5 + 12 = -0.25 \][/tex]
So, the turning point is at:
[tex]\[ (-3.5, -0.25) \][/tex]
### Summary of Answers:
1. The [tex]\( y \)[/tex]-intercept is [tex]\( 12 \)[/tex].
2. The [tex]\( x \)[/tex]-intercepts are [tex]\( -3 \)[/tex] and [tex]\( -4 \)[/tex].
3. The turning point is at [tex]\( (-3.5, -0.25) \)[/tex].
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