Description
A quadratic function is given.
f(x) = |
(b) Find its vertex and its x- and y-intercept(s). (If an answer does not exist, enter DNE.)
vertex | (x, y) = | |||
x-intercepts | (x, y) = | |||
(x, y) = | ||||
y-intercept | (x, y) = |
(c) Sketch its graph.
Explanation & Answer
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The given quadratic equation is f(x) = 2x^2 + 6x
To find the standard form, factorize the equation as x(2x+6)=0 Now, use the MPZ (Multiplication Property of Zero) to say that x=0 or 2x+6=0 Which gives you two solutions: x=0 or x=−3 To find the x-intercepts, we need to solve equation 2x^2+6x=0 y - intercept is point:Y(0,0) To compute y - coordinate of y - intercept, we will find f(0). In this example we have: f(0)=2⋅02+6⋅0+=0 Vertex is point: V=(−32,−92) Explanation: to find the x - coordinate of the vertex we use formula x=−b/2a In this example: a=2,b=6,c=0. So, the x-coordinate of the vertex is: x=−b/2a=−62⋅2=−32 f(−32)=2(−32)2+6⋅(−32)+0=−92 Focus is point: F=(−32,−358) Explanation: The x - coordinate of the focus is x=−b/2a The y - coordinate of the focus is y=1−b^2/4a +C