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Answer:
Graphing Quadratic Equations Using the Axis of Symmetry
A quadratic equation is a polynomial equation of degree 22 . The standard form of a quadratic equation is
0=ax2+bx+c0=ax2+bx+c
where a,ba,b and cc are all real numbers and a≠0a≠0 .
If we replace 00 with yy , then we get a quadratic function
y=ax2+bx+cy=ax2+bx+c
whose graph will be a parabola .
The axis of symmetry of this parabola will be the line x=−b2ax=−b2a . The axis of symmetry passes through the vertex, and therefore the xx -coordinate of the vertex is −b2a−b2a . Substitute x=−b2ax=−b2a in the equation to find the yy -coordinate of the vertex. Substitute few more xx -values in the equation to get the corresponding yy -values and plot the points. Join them and extend the parabola.
Example 1:
Graph the parabola y=x2−7x+2y=x2−7x+2 .
Compare the equation with y=ax2+bx+cy=ax2+bx+c to find the values of aa , bb , and cc .
Here, a=1,b=−7a=1,b=−7 and c=2c=2 .
Use the values of the coefficients to write the equation of axis of symmetry .
The graph of a quadratic equation in the form y=ax2+bx+cy=ax2+bx+c has as its axis of symmetry the line x=−b2ax=−b2a . So, the equation of the axis of symmetry of the given parabola is x=−(−7)2(1)x=−(−7)2(1) or x=72x=72 .
Substitute x=72x=72 in the equation to find the yy -coordinate of the vertex.
y=(72)2−7(72)+2 =494−492+2 =49 − 98 + 84 =−414y=(72)2−7(72)+2 =494−492+2 =49 − 98 + 84 =−414
Therefore, the coordinates of the vertex are (72,−414)(72,−414) .
Now, substitute a few more xx -values in the equation to get the corresponding yy -values.
Step-by-step explanation:
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