Solving Quadratic Equations by Graphing

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Exercises 25 There are three steps to solving a quadratic equation by graphing.Write the equation in standard form, ax2+bx+c=0. Graph the related function y=ax2+bx+c. Find the x-intercepts of the graph, if any.The solutions of ax2+bx+c=0 are the x-intercepts of the graph of y=ax2+bx+c. Since the given equation, x2+3x=18, is not written in standard form, we have to rewrite it. x2+3x=18⇔x2+3x−18=0​ Now we can graph the related function y=x2+3x−18. Note that the function on the given graph is y=x2+3x. The error is that the equation was not written in standard form before determining the related function and graphing it. Let's graph the correct related function and determine its x-intercepts.The solutions of the equation are x=-6 and x=3.
Exercises 26 There are three steps to solving a quadratic equation by graphing.Write the equation in standard form, ax2+bx+c=0. Graph the related function y=ax2+bx+c. Find the x-intercepts of the graph, if any.The solutions of ax2+bx+c=0 are the x-intercepts of the graph of y=ax2+bx+c. Since the given equation, x2+6x+9=0, is already written in standard form, we can determine the function related to it. Equation: Related Function: ​x2+6x+9=0y=x2+6x+9​ Now we can graph the related function and determine its x-intercepts.The graph has one x-intercept, x=-3. Therefore, the solution of the equation is x=-3. The graph given in the exercise is identical to ours. Then why the given answer is wrong? It is because they have wrongfully used the y-intercept of the graph as the solution.Remember that the solutions of a quadratic equation are always the x-intercepts of the graph of the related function!
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Exercises 29 We will solve the equation by graphing each side of the equation. The x-coordinates of the points of intersection of the graphs are the solutions of the equation. x2=10−3x⇒ LHS:y=x2RHS:y=10−3x​​ Let's graph the functions related to the left-hand side and the right-hand side of the equation one at a time.Left-Hand Side We will start by graphing the function y=x2. It is the parent quadratic function. The vertex of the graph of y=x2 is (0,0) and its axis of symmetry is x=0, or the y-axis. Next let's make a table of values to help us graphing.xy=x2 11 39 525 Let's plot these point and their reflection across the axis of symmetry.Finally we will connect the points with a smooth curve.Right-Hand Side Now we will graph the linear function y=10−3x. Let's rewrite the equation of the second function in slope-intercept form. We will also highlight the slope m and the y-intercept b. y=10−3x⇔y=-3x+10​ We will plot the y-intercept and use the slope to plot another point. Then we will connect the points with a straight edge.Connecting the Graphs The last step is connecting the graphs and determining their points of intersection. Recall that the x-coordinates of the points of intersection are the solutions of the given equation.The points of intersection of the graphs are (-5,25) and (2,4). Therefore, x=-5 and x=2 are the solutions of the equation x2=10−3x.
Exercises 30 We will solve the equation by graphing each side of the equation. The x-coordinates of the points of intersection of the graphs are the solutions of the equation. 2x−3=x2⇒ LHS:y=2x−3RHS:y=x2​​ Let's graph the functions related to the left-hand side and the right-hand side of the equation one at a time.Left-Hand Side We will start by graphing the linear function y=2x−3. Let's rewrite the equation of the first function in slope-intercept form. We will also highlight the slope m and the y-intercept b. y=2x−3⇔y=2x+(-3)​ We will plot the y-intercept and use the slope to plot another point. Then we will connect the points with a straight edge.Right-Hand Side Now we will graph the function y=x2. It is the parent quadratic function. The vertex of the graph of y=x2 is (0,0) and its axis of symmetry is x=0, or the y-axis. Next let's make a table of values to help us graphing.xy=x2 11 24 Let's plot these point and their reflection across the axis of symmetry.Finally we will connect the points with a smooth curve.Connecting the Graphs The last step is connecting the graphs and determining their points of intersection. Recall that the x-coordinates of the points of intersection are the solutions of the given equation.The graphs do not intersect. Therefore, the equation 2x−3=x2 has no real solutions.
Exercises 31 We will solve the equation by graphing each side of the equation. The x-coordinates of the points of intersection of the graphs are the solutions of the equation. 5x−7=x2⇒ LHS:y=5x−7RHS:y=x2​​ Let's graph the functions related to the left-hand side and the right-hand side of the equation one at a time.Left-Hand Side We will start by graphing the linear function y=5x−7. Let's rewrite the equation of the first function in slope-intercept form. We will also highlight the slope m and the y-intercept b. y=5x−7⇔y=5x+(-7)​ We will plot the y-intercept and use the slope to plot another point. Then we will connect the points with a straight edge.Right-Hand Side Now we will graph the function y=x2. It is the parent quadratic function. The vertex of the graph of y=x2 is (0,0) and its axis of symmetry is x=0, or the y-axis. Next let's make a table of values to help us graphing.xy=x2 11 24 39 Let's plot these point and their reflection across the axis of symmetry.Finally we will connect the points with a smooth curve.Connecting the Graphs The last step is connecting the graphs and determining their points of intersection. Recall that the x-coordinates of the points of intersection are the solutions of the given equation.The graphs do not intersect. Therefore, the equation 5x−7=x2 has no real solutions.
Exercises 32 We will solve the equation by graphing each side of the equation. The x-coordinates of the points of intersection of the graphs are the solutions of the equation. x2=6x−5⇒ LHS:y=x2RHS:y=6x−5​​ Let's graph the functions related to the left-hand side and the right-hand side of the equation one at a time.Left-Hand Side We will start by graphing the function y=x2. It is the parent quadratic function. The vertex of the graph of y=x2 is (0,0) and its axis of symmetry is x=0, or the y-axis. Next let's make a table of values to help us graphing.xy=x2 11 39 525 Let's plot these point and their reflection across the axis of symmetry.Finally we will connect the points with a smooth curve.Right-Hand Side Now we will graph the linear function y=6x−5. Let's rewrite the equation of the second function in slope-intercept form. We will also highlight the slope m and the y-intercept b. y=6x−5⇔y=6x+(-5)​ We will plot the y-intercept and use the slope to plot another point. Then we will connect the points with a straight edge.Connecting the Graphs The last step is connecting the graphs and determining their points of intersection. Recall that the x-coordinates of the points of intersection are the solutions of the given equation.The points of intersection of the graphs are (1,1) and (5,25). Therefore, x=1 and x=5 are the solutions of the equation x2=6x−5.
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