Answer:
[tex]y=68x-235[/tex]
Step-by-step explanation:
[tex]f(x) = 5x^2-2x + 10[/tex]
Differentiate:
[tex]\implies f'(x)=10x-2[/tex]
Substitute [tex]x=7[/tex] into [tex]f'(x)[/tex]:
[tex]\implies f'(7)=10(7)-2=68[/tex]
Therefore, 68 is the gradient of the tangent line at x=7:
[tex]\implies y-y_1=68(x-x_1)[/tex]
when [tex]x=7[/tex], [tex]y=5(7)^2-2(7)+10=241[/tex]
[tex]\implies y-241=68(x-7)[/tex]
[tex]\implies y=68x-235[/tex]
