Answer:
10
Step-by-step explanation:
Vertex
The x-coordinate of the vertex of a quadratic equation in the form
[tex]f(x)=ax^2+bx+c\quad \textsf{is} \quad -\dfrac{b}{2a}[/tex]
Given function:
[tex]f(x)=5x^2+20x-7[/tex]
[tex]\implies a=5, \quad b=20, \quad c=-7[/tex]
x-coordinate of the vertex
[tex]\implies -\dfrac{b}{2a}=-\dfrac{20}{2(5)}=-2[/tex]
To find the y-coordinate of the vertex, substitute the found value of x into the function:
[tex]\begin{aligned}\implies f(-2) & =5(-2)^2+20(-2)-7\\& = 5(4)-40-7\\& = 20-47\\& = -27\end{aligned}[/tex]
Therefore, the coordinates of the vertex are (-2, -27).
y-intercept
The y-intercept is when the curve crosses the y-axis, so when x = 0.
To find the y-coordinate of the y-intercept, substitute x = 0 into the function:
[tex]\begin{aligned}\implies f(0) & =5(0)^2+20(0)-7\\& = 0 + 0-7\\& = -7\end{aligned}[/tex]
Therefore, the coordinates of the y-intercept are (0, -7).
Slope
To find the slope of the line passing through the vertex and the y-intercept, simply substitute the found points into the slope formula:
[tex]\implies \sf slope=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{-27-(-7)}{-2-0}=\dfrac{-20}{-2}=10[/tex]
Therefore, the slope of the line passing through the vertex and the y-intercept of the given quadratic function is 10.
Learn more about slopes here:
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