Answer:
[tex]a=-12[/tex] and [tex]b=43[/tex]
Step-by-step explanation:
To find the constants, we need to replace the given point [tex](6;7)[/tex] into the given quadratic function: [tex]f(x)=x^{2}+ax+b[/tex]
So, from the problem we have [tex]x=6[/tex] and [tex]y=7[/tex]. Then:
[tex]f(x)=x^{2}+ax+b\\7=(6)^{2}+a(6)+b\\7=36+6a+b\\7-36=6a+b\\-29=6a+b[/tex]
However, if we have a minimum at [tex](6;7)[/tex], the first derivate of the function must be zero at 6.
[tex]f'(x)=2x+a=\\0=2(6)+a\\-12=a[/tex]
Now, we substitute this value in the previous expression:
[tex]-29=6a+b\\-29=6(-12)+b\\-29+72=b\\b=43[/tex]
Therefore, the values of the constants are [tex]a=-12[/tex] and [tex]b=43[/tex]
