Solving Quadratics By Sketching

Things to note before you start: [highlight]Plotting a quadratic equation on a graph will [b]always[/b] result in a curve[/highlight] [highlight]The points which cross the [b]x-axis[/b] are the root(s)[/highlight] [highlight]If there is one point touching the [b]x-axis[/b], there is one root[/highlight] [highlight]If there are no intersections between the curve and the axis there are no [u]real[/u] root(s)[/highlight] [highlight]Always make a values table before you actually sketch a graph (read on to see an example)[/highlight] [highlight]You may be specified a range in the form of: [code]number [lte] x [lte] number[/code] Which means make sure to make a table of values of [b]x[/b] based just in that range. [u]And if a range isn't specified, do a range of +3 to -3[/u][/highlight] [code]Before you start doing anything you need to make sure that one side [b]0[/b][/code] [code]Make sure that if a question provides an altered version of an equation in a sub question (i.e 2.b), that the side with [b]x[/b] on it, is identical to the original equation's [b]x[/b] side i.e: [b]1)[/b] [highlight]Original = x[squared]- 4x + 2 = 0 Altered = x[squared] - 4x - 4 = 0 You correct it to: [code]x[squared] - 4x + 2 = 6 (by adding 6)[/code] [/highlight] [b]2)[/b] [highlight]Original = x[squared]+ 3x - 2 = 0 Altered = x[squared]+ 3x + 2 = 0 You correct it to: [code]x[squared]+ 3x - 2 = -4 (by subtracting 4)[/code] [/highlight][u]Also remember that the value on the opposite side of the equation, is [b]y[/b][/u] [/code] [h3]Example 1[/h3] Say you have the quadratic equation [b]y=x[squared]-4x+2[/b]. And you need to plot it on a graph. The first thing you need to do is make a table. In the top row, plot [b]x[/b]. Now fill in the row underneath [b](the 'y' row)[/b], by substituting the [b]x[/b] with the corresponding [i]x[/i] values in the previous row. [highlight][b]i.e[/b]: [i]x=1[/i], [u]y=([b]1[squared][/b]) - (4 * [b]1[/b]) + 2[/u] which equals -1.[/highlight] It should look similar to this once done. [img src='https://res.cloudinary.com/deylrqt2d/image/upload/c_scale,h_230/v1484481451/quadratic_graph_table_mq1xe6.jpg'] Then make a graph of suitable size. [img src='http://res.cloudinary.com/deylrqt2d/image/upload/v1484442066/graph1_pre_kwu5ax.jpg'] Then plot the [b]x[/b] and [b]y[/b] points from the table on to the graph. [img src='http://res.cloudinary.com/deylrqt2d/image/upload/v1484481155/graph1_dfapin.jpg'] Now read the points off of the graph, where the curve crosses the [b]x[/b] axis. [img src='https://res.cloudinary.com/deylrqt2d/image/upload/v1484489138/graph1_final_k0gvqw.jpg'] The points that cross the [b]x-axis[/b] are [the roots]: [br] [code][u]x [approx] 3.35[/u][br] or[br] [u]x [approx] 0.5[/u] [/code] Therefore this curve has 2 distinct roots. [hr][h3]Example 2[/h3] [b]a)[/b]Sketch the graph [code]y = x[squared]- 2x - 4, -2 [lte] x [lte] 4 [the range is specified][/code] Values Table:[img src='http://res.cloudinary.com/deylrqt2d/image/upload/c_scale,h_230/v1484504172/quadratic_graph_table2_zxj7hf.jpg'] Plotted Graph:[img src='https://res.cloudinary.com/deylrqt2d/image/upload/v1484504794/graph2_final_x4qquu.gif'] [b]b) i.[/b] Use your graph to estimate the solution to: [code]x[squared] - 2x - 4 = 0 (remember the 0 represents [b]y[/b])[/code][This means it's asking you to find the values of [b]x[/b] where the curve goes over the [b]y = 0[/b] line] [b]Read them off of your graph[/b] [code]In this case the roots are:[br] [u]x [approx] 1.2[/u][br] or[br] [u]x [approx] 3.2[/u][/code] [hr] [b]b) ii.[/b] Use your graph to estimate the solution to: [code]x[squared] - 2x - 4 = 2 (remember the [u]2[/u] represents [b]y[/b])[/code][b]Read them off of your graph by drawing a straight line across [u]y = 2[/u] and, then estimating what the values of [u]x[/u] are where the quadratic curve intersects the line you drew[/b] [code]In this case the roots are:[br] [u]x [approx] 3.7[/u][br] or[br] [u]x [approx] -1.8[/u][/code] [hr] [b]b) iii.[/b] Use your graph to estimate the solution to: [code]x[squared] - 2x + 1 = 0 [br] (remember, you need to make the side with the [b]x[squared][/b] is identical to the original equation's [b]x[squared] side[/b] so you [b]subtract 5[/b]) [/code]So the equation you will estimate the solution to will be: [code]x[squared] - 2x - 4 = -5 [[b]y = -5[/b]] [/code][b]Read them off of your graph by drawing a straight line across [u]y = -5[/u] and, then estimating what the values of [u]x[/u] are where the quadratic curve intersects the line you drew[/b] [code]In this case there is only one root:[br] [u]x [approx] 1[/u][/code] [hr] [h3]Example 3[/h3] [b]a)[/b]Sketch the graph: [code]y = x[squared] - 3x - 1, -2 [lte] x [lte] 4[/code] Table of values: [img src='https://res.cloudinary.com/deylrqt2d/image/upload/c_scale,h_230/v1484513003/quadratic_graph_table3_kf92qi.gif'] Plotted Graph: [img src='https://res.cloudinary.com/deylrqt2d/image/upload/c_scale,h_855/v1484513869/graph3_final_rkayiy.gif'] [b]b)[/b]Use your graph to solve:[br] [code]x[squared] - 4x + 1 = 0[/code] [highlight]As we can see the [b]x[squared][/b] side of the equation above is different to the original's [b]x[squared][/b] side. So we can [b]add x[/b] and [b]subtract 2[/b] to get the original.[/highlight] So we get:[br] [code]x[squared] - 3x - 1 = x - 2[/code][br] This means [b]y = x - 2[/b] [highlight]We know that [b]x - 2[/b] is a line, so [b]y = x - 2[/b] is also a line.[br] We can work out how to plot this by first removing [b]x[/b] from the equation, so we plot the point [b] y = -2[/b] on the graph.[br] Then we remove [b]y[/b] from the equations and isolate [b]x[/b] by adding [b]2[/b] to both sides, to get [b]x = 2[/b], which we again plot on the graph. [br] Then we draw a straight line through the 2 plotted points, and get something like this: [img src='http://res.cloudinary.com/deylrqt2d/image/upload/c_scale,h_712/v1484514740/graph4_extra_ymmng0.gif'] [/highlight] Then note the intersection(s) on the line by the quadratic curve. [code] The roots are:[br] [u]x [approx] 0.25[/u][br] or[br] [u]x [approx] 3.8[/u] [/code] [hr]