Which equation represents y = −x2 + 6x + 7 in vertex form? y = −(x + 3)2 + 4 y = −(x − 3)2 + 10 y = −(x + 3)2 + 2 y = −(x − 3)2

+ 16

Mathematics

1 answer:

Colt1911 [192]3 years ago

8 0

$\text{The vertex form:}\ y=a(x-h)^2+k\\\\f(x)=ax^2+bx+c\to h=\dfrac{-b}{2a},\ k=f(h)\\\\\text{We have}\ y=-x^2+6x+7\to f(x)=-x^2+6x+7\\\\a=-1,\ b=6,\ c=7\\\\h=\dfrac{-6}{2(-1)}=\dfrac{-6}{-2}=3\\\\k=f(3)=-3^2+6(3)+7=-9+18+7=16\\\\Answer:\ \boxed{y=-(x-3)^2+16}$

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The given line segment has a midpoint at (3, 1) and goes through (2, 4), (3, 1), and (4, -2). We can use any two of the three points to calculate the equation of the line. Let us use the points (2, 4) and (4, -2)

Therefore the line goes through (2, 4) and (4, -2). The equation of a line passing through $(x_1,y_1)\ and\ (x_2,y_2)$ is:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Therefore the line passing through (2, 4) and (4, -2) has an equation:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\\frac{y-4}{x-2}=\frac{-2-4}{4-2}\\\frac{y-4}{x-2}=\frac{-6}{2}\\y-4=x-2(-3)\\y-4=-3x+6\\y=-3x+10$

Comparing with the general equation of line: y = mx + c, the slope (m) = -3 and the intercept on the y axis (c) = 10

Two lines are said to be perpendicular if the product of their slope is -1. If the slope of line one is m1 and the slope of line 2 = m2, then the two lines are perpendicular if:

$m_1m_2=-1$ .

Therefore The slope (m2) of the perpendicular bisector of y = -3x + 10 is:

$m_1m_2=-1\\-3m_2=-1\\m_2=\frac{1}{3}$

Since it is the perpendicular bisector of the given line segment, it passes through the midpoint (3, 1). The equation of the perpendicular bisector is:

$\frac{y-y_1}{x-x_1}=m\\\frac{y-1}{x-3}=\frac{1}{3}\\ y-1= \frac{1}{3}(x-3)\\ y-1=\frac{1}{3}x-1\\y=\frac{1}{3}x$