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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). $$f(x)=x^{2}+3 x+2$$

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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 $$(-2, 0)$$.
See the step by step solution

## 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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