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Problem 1

Sketch the graphs of the quadratic functions, indicating the coordinates of the vertex, the y-intercept, and the \(x\) -intercepts (if any). $$ f(x)=x^{2}+3 x+2 $$

Short Answer

Expert verified
The graph of the quadratic function \(f(x) = x^2 + 3x + 2\) is a parabola with vertex at \((-1.5, 0.25)\), y-intercept at \((0, 2)\), and x-intercepts at \((-1, 0)\) and \((-2, 0)\).
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Step 1: Find the vertex of the parabola

The x-coordinate of the vertex can be found using the formula \( -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = 3 \), so substituting these values in we get \( -\frac{3}{2(1)} = -1.5 \). The y-coordinate of the vertex is found by substituting the x-coordinate of the vertex into the equation. \( f(-1.5) = (-1.5)^2 + 3(-1.5) + 2 = 0.25 \). So, the vertex of the parabola is \((-1.5, 0.25)\).

Step 2: Identify the y-intercept

The y-intercept of the function is the value of \( f \) at \( x = 0 \). So, \( f(0) = (0)^2 + 3(0) + 2 = 2 \). So, the y-intercept is \( (0, 2) \).

Step 3: Determine the x-intercepts

The x-intercepts of the parabola are the points where it intersects the x-axis. We can find these by setting \( f(x) = 0 \) and solving for \( x \). Setting \( f(x) = x^2 + 3x + 2 = 0 \) and using the quadratic formula \( x = \frac{ -b \pm \sqrt {b^2 - 4ac} }{2a} \) we get \( x = \frac{-3 \pm \sqrt{3^2 - 4(1)(2)}}{2(1)} = -1, -2 \). So, the x-intercepts of the parabola are \((-1,0)\) and \((-2,0)\). The graph of the quadratic function \( f(x) = x^2 + 3x + 2 \) will be a parabola with vertex at \((-1.5, 0.25)\), y-intercept at \( (0, 2) \), and x-intercepts at \((-1, 0)\) and \((-2, 0)\).

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