ow do you find a and also put in standard form what is the axis of symmetry and what is the y intercept find the zeros and is the parabola upward or downward

Question

F(x)= ( x-6.36)^2 +1.11  what is a and put it in standard form

how do you find a and also put in standard form what is the axis of symmetry and  what is the y intercept find the zeros  and is the parabola upward or downward

Answer 1

If by a you are referring to the form (y-k)=a(x-h)^2. So a=1 and (h,k) is the vertex, with x=h as the line of symmetry. The parabola is like a U, because a>0. (h,k) is (6.36,1.11) and the axis of symmetry is x=6.36. When x=0, y=f(x)=ah^2+k=6.36^2+1.11=41.5596, y intercept. The zeroes are not real, they're complex, because the graph does not cross the x axis (when y=f(x)=0). The minimum value of y is at the vertex.

The polynomial can also be written in the form y=ax^2+bx+c;

a=1, so y=f(x)=x^2+bx+c=x^2-12.72x+40.4496+1.11=x^2-12.72x+41.5596.

The complex zeroes are found by solving (x-6.36)=+sqrt(-1.11)=+isqrt(1.11); x=6.36+isqrt(1.11)=6.36+1.0536i.

Comments

So you say a=1, so y=f(x)=x^2+bx+c=x^2-12.72x+40.4496+1.11=x^2-12.72x+41.5596 and The complex zeroes are found by solving (x-6.36)=+sqrt(-1.11)=+isqrt(1.11); x=6.36+isqrt(1.11)=6.36+1.0536i. Good answer Rod! Awesome! You cool.

Answer 2

You might go with a=1 and (h,k) is the vertx, with x=h as the lin of symmetry like Rod said. GO with his instead of mine because he is more specific than I am.