1. Parabola, 3, 1, -1: y=2(x+3)-4x+(x-3)(x+1) simplifies to y=x^2-4x+3, and the zeros of the graph of the equation occur when y=0. To solve 0=x2-4x+3, use the factored form 0=(x-3)(x-1) and set each factor equal to zero. Then, solve for x: (x-3)(x-1)=0 (x-3)=0 (x-1)=0 x=3 x=1 The graph of the quadratic equation ax^2+bx+c is a parabola with an axis of symmetry x=-b/2a, and the maximum or minimum value of a parabola lies on the axis of symmetry. Find the equation which represents the axis of symmetry, and find the y-value for the determined value of x: For =x2-4x+3, a=1,b=-4,c=3. The axis of symmetry is x= (-(-4))/2(1) =2. When x=2, y=(2)2-4(2)+3=4-8+3=-1.
2. C, D, E; y=3/2 x+3: The rate of change in a linear relationship is constant. In the given table, the output, or y, value decreases by three as the input, or x, value decreases by 2. The slope m of the line between the two given points is, therefore, (-3)/(-2)=3/2. A linear equation can be written in slope-intercept form, y=mx+b. Substitute one of the given x-y pairs and the determined slope into the equation to find the y-intercept b: y=mx+b 3=3/2(0)+b 3=b Using the determined slope and y-intercept, the equation of the line in slope-intercept form is y=3/2 x+3. Using this equation, input values x=0,4,-2,-5 gives output values y=3,9,0,-4.5, respectively. Of the answer choices, (-2,0),(4,9), and (-5,-4.5) lie on the same line the one which passes through given points (2,6) and (0,3).
The graph as x approaches -2 from the left is linear; the slope of the line is 2, and, if the line were extended, it would cross the y-axis at 4. Therefore, in slope-intercept form, the equation of the line is y=2x+4. The function is defined by this line for all x-values less than or equal to -2. Once x exceeds -2, the graph is a parabola. This parabola is the graph of y=x2 shifted down one unit, so its equation is y=x2-1.
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