How to solve quadratic equations by graphing ??

Mathematics

2 answers:

Inga [223]3 years ago

8 0

Answer:

Well if we were to graph this quadratic equation. We would be looking for places on that graph where Y is equal to zero. Well those are places where the graph crosses the x axis.

Step-by-step explanation:

zaharov [31]3 years ago

7 0

Answer:

find points
plot them
draw a curve through them
identify the x-intercepts

Step-by-step explanation:

You graph a quadratic the way you graph any equation.

First you make a table of points.
Then you plot those points on a graph.
Finally, you draw a smooth curve through the points you have plotted.

The solution is generally where the graph crosses the x-axis, depending on what you're graphing.

We assume you want to graph an equation of the form ...

y = ax^2 +bx +c

Often, we are interested in the values of x that make y = 0. Sometimes, we want other solutions, for example, the one(s) that make y = 194. That is why I say "depending on what you're graphing" in the text above.

For a quadratic, it is helpful to start at the line of symmetry. The line of symmetry for this equation is the vertical line at ...

x = -b/(2a)

The graph will be symmetrical about this line. Points on either side of this line will be the same distance from it.

So, once you find the x-value for this line of symmetry, you evaluate the quadratic for that value of x. The resulting y-value will be the y-value of the vertex, the most extreme point on the curve. If a>0, it will be the lowest point on the curve. The curve will open upward from there. If a < 0, it will be the highest point on the curve, which will open downward.

Another point you can plot is the y-intercept. The value of that is 'c' in the above equation, so (0, c) is a point on the curve. Of course, another point on the curve is found where this one is reflected across the line of symmetry.

Personally, I usually find it convenient to choose values of x that differ from the line of symmetry by 1, 2, 3, or some other small integer value. You will find for these values of x, the corresponding y-values differ from the vertex by a, 4a, 9a. Of course, for some quadratics, it is appropriate to choose a different set of points than these.

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Worked-out Example

Consider ...

y = 2x^2 -x -6

The values of a, b, c are: a=2, b=-1, c=-6.

One point you can graph right away is the y-intercept: (0, -6).

The line of symmetry is ...

x = -b/(2a) = -(-1)/(2(2)) = 1/4

The corresponding value of y is ...

y = 2(1/4)^2 -1(1/4) -6 = 2/16 -1/4 -6 = -(6 1/8)

So, the vertex of this graph is (1/4, -6 1/8).

As we noted above, we can add points at ±1, ±2, ±3 from the vertex fairly easily.

For x = 1/4 ±1, y will be -6 1/8 +2(1^2) = -4 1/8. So, two more points are ...

(1 1/4, -4 1/8) and (-3/4, -4 1/8)

For x = 1/4 ±2, y will be -6 1/8 +2(2^2) = 1 7/8. So, two more points are ...

(2 1/4, 1 7/8) and (-1 3/4, 1 7/8)

You will notice that the graph crosses the x-axis between -3/4 and -1 3/4, and also between 1 1/4 and 2 1/4. You might reasonably guess that values of x of 2 or -1 1/2 would give y-values of zero. You would be right.

For x=2, we have ...

y = 2(2^2) -(2) -6 = 8 -2 -6 = 0

For x = -1.5, we have ...

y = 2(-1.5)^2 -(-1.5) -6 = 2(2.25) +1.5 -6 = 4.5 +1.5 -6 = 0

So, the solutions are also points on the graph:

(2, 0) and (-1.5, 0)

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Additional comment

Graphing calculator apps are available for smartphones, tablets, and desktop computers. I personally like to use one of those for making graphs of equations. One that is reasonably easy to learn is shown in the attachment. Desmos provides this one.

The attachment shows part of the table we computed, and a plot of those points and the curve through them. The x-intercepts are also shown.