Let w =Vector O R where O(–5, 6) and R(4, –2). What is the component form of One-fifthw?
2 answers:
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Answer:
Vertex: (-4, -1)
Y-intercept: (0, 15)
Step-by-step explanation:
Given the quaratic function, h(x) = x² + 8x + 15:
In order to determine the vertex of the given function, we can use the formula, .
In the quadratic function, h(x) = x² + 8x + 15, where:
a = 1, b = 8, and c = 15:
Substitute the given values for <em>a</em> and <em>b</em> into the equation to solve for the x-coordinate of the vertex.
x = -4
Subsitute the value of the x-coordinate into the given function to solve for the <u>y-coordinate of the vertex</u>:
h(x) = x² + 8x + 15
h(-4) = (-4)² + 8(-4) + 15
h(-4) = 16 - 32 + 15
h(-4) = -1
Therefore, the vertex of the given function is (-4, -1).
<h3>Solve for the Y-intercept:</h3>The <u>y-intercept</u> is the point on the graph where it crosse the y-axis. In order to find the y-intercept of the function, set x = 0, and solve for the y-intercept:
h(x) = x² + 8x + 15
h(0) = (0)² + 8(0) + 15
h(0) = 0 + 0 + 15
h(0) = 15
Therefore, the y-intercept of the quadratic function is (0, 15).