the quadratic function f(x)=ax2+bx+5 is given. Determine a and b so that the graph of the function is symmetrical about the ordinate axis and passes through the point T(2,7)

If the graph should be symmetrical then x=2,y=7 and x=-2 y=7 if x=2 y=7 $$ \left(2\right)^2a+\left(2\right)b+5=7 $$ $$ \left(4\right)a+\left(2\right)b=2 $$ if x=-2 y=7 $$ \left(4\right)a+\left(-2\right)b=2 $$ Solve for a an b $$ a=\frac{1}{2},b=0 $$ Answer: $$ f\left(x\right)=\frac{1}{2}x^2+5 $$