Question

SKETCHING GRAPHS Sketch the graph of the function. Label the vertex. $$ y=2 x^{2}+6 x-5 $$

Short answer

The vertex of the function \(y=2 x^{2}+6 x-5\) is at \((-1.5, -5.5)\). Sketching this on a graph results in a parabola with the vertex at \((-1.5,-5.5)\), opening upwards and crossing the y-axis at \(y = -5\).

Step-by-step solution

Determine the vertex

First, write down the quadratic function. Here: \( y = 2x^2 + 6x - 5 \). For a function of the form \( y = ax^2 + bx + c \), the vertex can be found using the formula \(h = -b/2a\). In this case, \(a = 2, b = 6\). Thus, the x-coordinate of the vertex, \(h\), is \(-6/(2*2) = -1.5\). To find the y-coordinate, \(k\), substitute \(h\) into the equation: \( y = 2(-1.5)^2 + 6(-1.5) - 5 = -5.5 \). Therefore, the vertex is \((-1.5, -5.5)\).

Plot the vertex and the parabola

Now that the vertex has been found, you can plot this point on a coordinate plane. Given this is a parabola, it’s known that it will be symmetric about the line \(x = h\), and it will open upwards because \(a > 0\). That means the parabola will be lowest at the vertex. The y-intercept can also be found by setting \(x = 0\) in the original equation, which results in \(y = -5\). You have two points now: the vertex and the y-intercept. Plot these points, sketch the axis of symmetry at \(x = h\), and complete the graph using the symmetry of the parabola.