Question

The parabola starts at left parenthesis 1 comma 0 right parenthesis and passes through left parenthesis 2 comma 5 right parenthesis to its vertex at left parenthesis 5 comma 10 right parenthesis. It continues through left parenthesis 8 comma 5 right parenthesis and ends at left parenthesis 9 comma 0 right parenthesis. All values are approximate.


The graph shows the trajectory of a firework. Which equation in vertex form matches this graph?

(1 point)
Responses

y=−12(x−5)2−10
y equals negative Start Fraction 1 over 2 End Fraction left parenthesis x minus 5 right parenthesis squared minus 10

y=−12(x−5)2+10
y equals negative Start Fraction 1 over 2 End Fraction left parenthesis x minus 5 right parenthesis squared plus 10

y=12(x−5)2+10
y equals Start Fraction 1 over 2 End Fraction left parenthesis x minus 5 right parenthesis squared plus 10

y=12(x+5)2+10
y equals Start Fraction 1 over 2 End Fraction left parenthesis x plus 5 right parenthesis squared plus 10

Answer 1

y = −(1/2)(x − 5)^2 + 10

Reason: The vertex form is y = a(x−5)^2 + 10 (vertex (5,10)). The parabola opens downward, so a < 0; the only downward choice with +10 is −(1/2)(x−5)^2 + 10 (which gives values close to the plotted points).