Question

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

Answer 1

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). 
$0=(2)^2+2b+c\\-4=2b+c$

Second, (0, 14).
$14=(0)^2+0b+c\\c=14$

So we have a value (14) for c. We can substitute that into our first equation to find b.

$-4=2b+c\\-4=2b+14\\-18=2b\\b=-9$

We can now plug in our values for b and c into the equation to get its standard form.

$y=x^2-9x+14$

To find the vertex, we can convert this equation to vertex form $y=a(x-h)^2+k$ by completing the square.

$y=x^2-9x+14$

$y=x^2-9x+(-4.5)^2+14-(-4.5)^2$

$y=x^2-9x+20.25+14-20.25$

$y=(x-4.5)^2-6.25$

Thus, the vertex is (4.5, –6.25).

We can confirm the solution graphically (see attachment).