here is a sketch of y=x^2 + bx + c. the curve intersects the x-axis at (2,0) and point P. the y-axis at (0,14) work out the x-coordinate of the turning point of the graph
Accepted Solution
A:
To find the turning point, or vertex, of this parabola, we need to work out the values of the coefficients b and c.
We are given two different solutions of the equation.
First, (2, 0). [tex]0=(2)^2+2b+c\\-4=2b+c[/tex]
Second, (0, 14). [tex]14=(0)^2+0b+c\\c=14[/tex]
So we have a value (14) for c. We can substitute that into our first equation to find b.
[tex]-4=2b+c\\-4=2b+14\\-18=2b\\b=-9[/tex]
We can now plug in our values for b and c into the equation to get its standard form.
[tex]y=x^2-9x+14[/tex]
To find the vertex, we can convert this equation to vertex form [tex]y=a(x-h)^2+k[/tex] by completing the square.
[tex]y=x^2-9x+14[/tex]
[tex]y=x^2-9x+(-4.5)^2+14-(-4.5)^2[/tex]
[tex]y=x^2-9x+20.25+14-20.25[/tex]
[tex]y=(x-4.5)^2-6.25[/tex]
Thus, the vertex is (4.5, –6.25).
We can confirm the solution graphically (see attachment).